If we list all the natural numbers below $10$ that are multiples of $3$ or $5$, we get $3, 5, 6$ and $9$. The sum of these multiples is $23$.

Find the sum of all the multiples of $3$ or $5$ below $1000$.

Answer: 233168


Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with $1$ and $2$, the first $10$ terms will be:
$$1, 2, 3, 5, 8, 13, 21, 34, 55, 89, \dots$$

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Answer: 4613732


The prime factors of $13195$ are $5, 7, 13$ and $29$.

What is the largest prime factor of the number $600851475143$?

Answer: 6857


A palindromic number reads the same both ways. The largest palindrome made from the product of two $2$-digit numbers is $9009 = 91 \times 99$.

Find the largest palindrome made from the product of two $3$-digit numbers.

Answer: 906609


$2520$ is the smallest number that can be divided by each of the numbers from $1$ to $10$ without any remainder.

What is the smallest positive number that is evenly divisibledivisible with no remainder by all of the numbers from $1$ to $20$?

Answer: 232792560


The sum of the squares of the first ten natural numbers is,

$$1^2 + 2^2 + ... + 10^2 = 385.$$
The square of the sum of the first ten natural numbers is,

$$(1 + 2 + ... + 10)^2 = 55^2 = 3025.$$
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is $3025 - 385 = 2640$.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

Answer: 25164150


By listing the first six prime numbers: $2, 3, 5, 7, 11$, and $13$, we can see that the $6$th prime is $13$.

What is the $10\,001$st prime number?

Answer: 104743


Answer: 23514624000


A Pythagorean triplet is a set of three natural numbers, $a \lt b \lt c$, for which,
$$a^2 + b^2 = c^2.$$

For example, $3^2 + 4^2 = 9 + 16 = 25 = 5^2$.

There exists exactly one Pythagorean triplet for which $a + b + c = 1000$.
Find the product $abc$.

Answer: 31875000


The sum of the primes below $10$ is $2 + 3 + 5 + 7 = 17$.

Find the sum of all the primes below two million.

Answer: 142913828922


Answer: 70600674


The sequence of triangle numbers is generated by adding the natural numbers. So the $7$th triangle number would be $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$. The first ten terms would be:
$$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \dots$$

Let us list the factors of the first seven triangle numbers:

$$\begin{align}
\mathbf 1 &\colon 1\\
\mathbf 3 &\colon 1,3\\
\mathbf 6 &\colon 1,2,3,6\\
\mathbf{10} &\colon 1,2,5,10\\
\mathbf{15} &\colon 1,3,5,15\\
\mathbf{21} &\colon 1,3,7,21\\
\mathbf{28} &\colon 1,2,4,7,14,28
\end{align}$$
We can see that $28$ is the first triangle number to have over five divisors.

What is the value of the first triangle number to have over five hundred divisors?

Answer: 76576500


Answer: 5537376230


The following iterative sequence is defined for the set of positive integers:

- $n \to n/2$ ($n$ is even)

- $n \to 3n + 1$ ($n$ is odd)

Using the rule above and starting with $13$, we generate the following sequence:
$$13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1.$$

It can be seen that this sequence (starting at $13$ and finishing at $1$) contains $10$ terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at $1$.

Which starting number, under one million, produces the longest chain?

NOTE: Once the chain starts the terms are allowed to go above one million.

Answer: 837799


Starting in the top left corner of a $2 \times 2$ grid, and only being able to move to the right and down, there are exactly $6$ routes to the bottom right corner.

How many such routes are there through a $20 \times 20$ grid?

Answer: 137846528820


$2^{15} = 32768$ and the sum of its digits is $3 + 2 + 7 + 6 + 8 = 26$.

What is the sum of the digits of the number $2^{1000}$?

Answer: 1366


If the numbers $1$ to $5$ are written out in words: one, two, three, four, five, then there are $3 + 3 + 5 + 4 + 4 = 19$ letters used in total.

If all the numbers from $1$ to $1000$ (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, $342$ (three hundred and forty-two) contains $23$ letters and $115$ (one hundred and fifteen) contains $20$ letters. The use of "and" when writing out numbers is in compliance with British usage.

Answer: 21124


By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is $23$.

3
7 4

2 4 6

8 5 9 3

That is, $3 + 7 + 4 + 9 = 23$.

Find the maximum total from top to bottom of the triangle below:

75

95 64

17 47 82

18 35 87 10

20 04 82 47 65

19 01 23 75 03 34

88 02 77 73 07 63 67

99 65 04 28 06 16 70 92

41 41 26 56 83 40 80 70 33

41 48 72 33 47 32 37 16 94 29

53 71 44 65 25 43 91 52 97 51 14

70 11 33 28 77 73 17 78 39 68 17 57

91 71 52 38 17 14 91 43 58 50 27 29 48

63 66 04 68 89 53 67 30 73 16 69 87 40 31

04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only $16384$ routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)

Answer: 1074


You are given the following information, but you may prefer to do some research for yourself.

- 1 Jan 1900 was a Monday.

- Thirty days has September,

April, June and November.

All the rest have thirty-one,

Saving February alone,

Which has twenty-eight, rain or shine.

And on leap years, twenty-nine.

- A leap year occurs on any year evenly divisible by 4, but not on a century unless it is divisible by 400.

How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?

Answer: 171


$n!$ means $n \times (n - 1) \times \cdots \times 3 \times 2 \times 1$.

For example, $10! = 10 \times 9 \times \cdots \times 3 \times 2 \times 1 = 3628800$,
and the sum of the digits in the number $10!$ is $3 + 6 + 2 + 8 + 8 + 0 + 0 = 27$.

Find the sum of the digits in the number $100!$.

Answer: 648


Let $d(n)$ be defined as the sum of proper divisors of $n$ (numbers less than $n$ which divide evenly into $n$).

If $d(a) = b$ and $d(b) = a$, where $a \ne b$, then $a$ and $b$ are an amicable pair and each of $a$ and $b$ are called amicable numbers.

For example, the proper divisors of $220$ are $1, 2, 4, 5, 10, 11, 20, 22, 44, 55$ and $110$; therefore $d(220) = 284$. The proper divisors of $284$ are $1, 2, 4, 71$ and $142$; so $d(284) = 220$.

Evaluate the sum of all the amicable numbers under $10000$.

Answer: 31626


Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alphabetical position in the list to obtain a name score.

For example, when the list is sorted into alphabetical order, COLIN, which is worth $3 + 15 + 12 + 9 + 14 = 53$, is the $938$th name in the list. So, COLIN would obtain a score of $938 \times 53 = 49714$.

What is the total of all the name scores in the file?

Answer: 871198282


A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of $28$ would be $1 + 2 + 4 + 7 + 14 = 28$, which means that $28$ is a perfect number.

A number $n$ is called deficient if the sum of its proper divisors is less than $n$ and it is called abundant if this sum exceeds $n$.

As $12$ is the smallest abundant number, $1 + 2 + 3 + 4 + 6 = 16$, the smallest number that can be written as the sum of two abundant numbers is $24$. By mathematical analysis, it can be shown that all integers greater than $28123$ can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

Answer: 4179871


A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are:

012   021   102   120   201   210

What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?

Answer: 2783915460


The Fibonacci sequence is defined by the recurrence relation:

$F_n = F_{n - 1} + F_{n - 2}$, where $F_1 = 1$ and $F_2 = 1$.
Hence the first $12$ terms will be:

$$\begin{align}
F_1 &= 1\\
F_2 &= 1\\
F_3 &= 2\\
F_4 &= 3\\
F_5 &= 5\\
F_6 &= 8\\
F_7 &= 13\\
F_8 &= 21\\
F_9 &= 34\\
F_{10} &= 55\\
F_{11} &= 89\\
F_{12} &= 144
\end{align}$$
The $12$th term, $F_{12}$, is the first term to contain three digits.

What is the index of the first term in the Fibonacci sequence to contain $1000$ digits?

Answer: 4782


A unit fraction contains $1$ in the numerator. The decimal representation of the unit fractions with denominators $2$ to $10$ are given:

$$\begin{align}
1/2 &= 0.5\\
1/3 &=0.(3)\\
1/4 &=0.25\\
1/5 &= 0.2\\
1/6 &= 0.1(6)\\
1/7 &= 0.(142857)\\
1/8 &= 0.125\\
1/9 &= 0.(1)\\
1/10 &= 0.1
\end{align}$$
Where $0.1(6)$ means $0.166666\cdots$, and has a $1$-digit recurring cycle. It can be seen that $1/7$ has a $6$-digit recurring cycle.

Find the value of $d \lt 1000$ for which $1/d$ contains the longest recurring cycle in its decimal fraction part.

Answer: 983


Euler discovered the remarkable quadratic formula:

$n^2 + n + 41$

It turns out that the formula will produce $40$ primes for the consecutive integer values $0 \le n \le 39$. However, when $n = 40, 40^2 + 40 + 41 = 40(40 + 1) + 41$ is divisible by $41$, and certainly when $n = 41, 41^2 + 41 + 41$ is clearly divisible by $41$.

The incredible formula $n^2 - 79n + 1601$ was discovered, which produces $80$ primes for the consecutive values $0 \le n \le 79$. The product of the coefficients, $-79$ and $1601$, is $-126479$.

Considering quadratics of the form:

$n^2 + an + b$, where $|a| < 1000$ and $|b| \le 1000$

where $|n|$ is the modulus/absolute value of $n$
e.g. $|11| = 11$ and $|-4| = 4$

Find the product of the coefficients, $a$ and $b$, for the quadratic expression that produces the maximum number of primes for consecutive values of $n$, starting with $n = 0$.

Answer: -59231


Starting with the number $1$ and moving to the right in a clockwise direction a $5$ by $5$ spiral is formed as follows:

21 22 23 24 25

20  7  8  9 10

19  6  1  2 11

18  5  4  3 12
17 16 15 14 13

It can be verified that the sum of the numbers on the diagonals is $101$.

What is the sum of the numbers on the diagonals in a $1001$ by $1001$ spiral formed in the same way?

Answer: 669171001


Answer: 9183


Surprisingly there are only three numbers that can be written as the sum of fourth powers of their digits:
$$\begin{align}
1634 &= 1^4 + 6^4 + 3^4 + 4^4\\
8208 &= 8^4 + 2^4 + 0^4 + 8^4\\
9474 &= 9^4 + 4^4 + 7^4 + 4^4
\end{align}$$

As $1 = 1^4$ is not a sum it is not included.

The sum of these numbers is $1634 + 8208 + 9474 = 19316$.

Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Answer: 443839


In the United Kingdom the currency is made up of pound (£) and pence (p). There are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p), and £2 (200p).
It is possible to make £2 in the following way:

1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p
How many different ways can £2 be made using any number of coins?

Answer: 73682


We shall say that an $n$-digit number is pandigital if it makes use of all the digits $1$ to $n$ exactly once; for example, the $5$-digit number, $15234$, is $1$ through $5$ pandigital.

The product $7254$ is unusual, as the identity, $39 \times 186 = 7254$, containing multiplicand, multiplier, and product is $1$ through $9$ pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a $1$ through $9$ pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

Answer: 45228


The fraction $49/98$ is a curious fraction, as an inexperienced mathematician in attempting to simplify it may incorrectly believe that $49/98 = 4/8$, which is correct, is obtained by cancelling the $9$s.

We shall consider fractions like, $30/50 = 3/5$, to be trivial examples.

There are exactly four non-trivial examples of this type of fraction, less than one in value, and containing two digits in the numerator and denominator.

If the product of these four fractions is given in its lowest common terms, find the value of the denominator.

Answer: 100


$145$ is a curious number, as $1! + 4! + 5! = 1 + 24 + 120 = 145$.

Find the sum of all numbers which are equal to the sum of the factorial of their digits.

Note: As $1! = 1$ and $2! = 2$ are not sums they are not included.

Answer: 40730


The number, $197$, is called a circular prime because all rotations of the digits: $197$, $971$, and $719$, are themselves prime.

There are thirteen such primes below $100$: $2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79$, and $97$.

How many circular primes are there below one million?

Answer: 55


The decimal number, $585 = 1001001001_2$ (binary), is palindromic in both bases.

Find the sum of all numbers, less than one million, which are palindromic in base $10$ and base $2$.

(Please note that the palindromic number, in either base, may not include leading zeros.)

Answer: 872187


The number $3797$ has an interesting property. Being prime itself, it is possible to continuously remove digits from left to right, and remain prime at each stage: $3797$, $797$, $97$, and $7$. Similarly we can work from right to left: $3797$, $379$, $37$, and $3$.

Find the sum of the only eleven primes that are both truncatable from left to right and right to left.

NOTE: $2$, $3$, $5$, and $7$ are not considered to be truncatable primes.

Answer: 748317


Take the number $192$ and multiply it by each of $1$, $2$, and $3$:

$$\begin{align}
192 \times 1 &= 192\\
192 \times 2 &= 384\\
192 \times 3 &= 576
\end{align}$$
By concatenating each product we get the $1$ to $9$ pandigital, $192384576$. We will call $192384576$ the concatenated product of $192$ and $(1,2,3)$.

The same can be achieved by starting with $9$ and multiplying by $1$, $2$, $3$, $4$, and $5$, giving the pandigital, $918273645$, which is the concatenated product of $9$ and $(1,2,3,4,5)$.

What is the largest $1$ to $9$ pandigital $9$-digit number that can be formed as the concatenated product of an integer with $(1,2, \dots, n)$ where $n \gt 1$?

Answer: 932718654


If $p$ is the perimeter of a right angle triangle with integral length sides, $\{a, b, c\}$, there are exactly three solutions for $p = 120$.

$\{20,48,52\}$, $\{24,45,51\}$, $\{30,40,50\}$

For which value of $p \le 1000$, is the number of solutions maximised?

Answer: 840


An irrational decimal fraction is created by concatenating the positive integers:
$$0.12345678910{\color{red}\mathbf 1}112131415161718192021\cdots$$

It can be seen that the $12$th digit of the fractional part is $1$.

If $d_n$ represents the $n$th digit of the fractional part, find the value of the following expression.
$$d_1 \times d_{10} \times d_{100} \times d_{1000} \times d_{10000} \times d_{100000} \times d_{1000000}$$

Answer: 210


We shall say that an $n$-digit number is pandigital if it makes use of all the digits $1$ to $n$ exactly once. For example, $2143$ is a $4$-digit pandigital and is also prime.

What is the largest $n$-digit pandigital prime that exists?

Answer: 7652413


The $n$th term of the sequence of triangle numbers is given by, $t_n = \frac12n(n+1)$; so the first ten triangle numbers are:
$$1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \dots$$

By converting each letter in a word to a number corresponding to its alphabetical position and adding these values we form a word value. For example, the word value for SKY is $19 + 11 + 25 = 55 = t_{10}$. If the word value is a triangle number then we shall call the word a triangle word.

Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, how many are triangle words?

Answer: 162


The number, $1406357289$, is a $0$ to $9$ pandigital number because it is made up of each of the digits $0$ to $9$ in some order, but it also has a rather interesting sub-string divisibility property.

Let $d_1$ be the $1$st digit, $d_2$ be the $2$nd digit, and so on. In this way, we note the following:

- $d_2d_3d_4=406$ is divisible by $2$

- $d_3d_4d_5=063$ is divisible by $3$

- $d_4d_5d_6=635$ is divisible by $5$

- $d_5d_6d_7=357$ is divisible by $7$

- $d_6d_7d_8=572$ is divisible by $11$

- $d_7d_8d_9=728$ is divisible by $13$

- $d_8d_9d_{10}=289$ is divisible by $17$

Find the sum of all $0$ to $9$ pandigital numbers with this property.

Answer: 16695334890


Pentagonal numbers are generated by the formula, $P_n=n(3n-1)/2$. The first ten pentagonal numbers are:
$$1, 5, 12, 22, 35, 51, 70, 92, 117, 145, \dots$$

It can be seen that $P_4 + P_7 = 22 + 70 = 92 = P_8$. However, their difference, $70 - 22 = 48$, is not pentagonal.

Find the pair of pentagonal numbers, $P_j$ and $P_k$, for which their sum and difference are pentagonal and $D = |P_k - P_j|$ is minimised; what is the value of $D$?

Answer: 5482660


Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

| Triangle |  | $T_n=n(n+1)/2$ |  | $1, 3, 6, 10, 15, \dots$ |
| Pentagonal |  | $P_n=n(3n - 1)/2$ |  | $1, 5, 12, 22, 35, \dots$ |
| Hexagonal |  | $H_n=n(2n - 1)$ |  | $1, 6, 15, 28, 45, \dots$ |

It can be verified that $T_{285} = P_{165} = H_{143} = 40755$.

Find the next triangle number that is also pentagonal and hexagonal.

Answer: 1533776805


It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.

$$\begin{align}
9 = 7 + 2 \times 1^2\\
15 = 7 + 2 \times 2^2\\
21 = 3 + 2 \times 3^2\\
25 = 7 + 2 \times 3^2\\
27 = 19 + 2 \times 2^2\\
33 = 31 + 2 \times 1^2
\end{align}$$
It turns out that the conjecture was false.

What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?

Answer: 5777


The first two consecutive numbers to have two distinct prime factors are:

$$\begin{align}
14 &= 2 \times 7\\
15 &= 3 \times 5.
\end{align}$$
The first three consecutive numbers to have three distinct prime factors are:

$$\begin{align}
644 &= 2^2 \times 7 \times 23\\
645 &= 3 \times 5 \times 43\\
646 &= 2 \times 17 \times 19.
\end{align}$$
Find the first four consecutive integers to have four distinct prime factors each. What is the first of these numbers?

Answer: 134043


Answer: 9110846700


The arithmetic sequence, $1487, 4817, 8147$, in which each of the terms increases by $3330$, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the $4$-digit numbers are permutations of one another.

There are no arithmetic sequences made up of three $1$-, $2$-, or $3$-digit primes, exhibiting this property, but there is one other $4$-digit increasing sequence.

What $12$-digit number do you form by concatenating the three terms in this sequence?

Answer: 296962999629


The prime $41$, can be written as the sum of six consecutive primes:

$$41 = 2 + 3 + 5 + 7 + 11 + 13.$$
This is the longest sum of consecutive primes that adds to a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime, contains $21$ terms, and is equal to $953$.

Which prime, below one-million, can be written as the sum of the most consecutive primes?

Answer: 997651


By replacing the 1st digit of the 2-digit number *3, it turns out that six of the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.

By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit number is the first example having seven primes among the ten generated numbers, yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993. Consequently 56003, being the first member of this family, is the smallest prime with this property.

Find the smallest prime which, by replacing part of the number (not necessarily adjacent digits) with the same digit, is part of an eight prime value family.

Answer: 121313


It can be seen that the number, $125874$, and its double, $251748$, contain exactly the same digits, but in a different order.

Find the smallest positive integer, $x$, such that $2x$, $3x$, $4x$, $5x$, and $6x$, contain the same digits.

Answer: 142857


There are exactly ten ways of selecting three from five, 12345:

123, 124, 125, 134, 135, 145, 234, 235, 245, and 345

In combinatorics, we use the notation, $\displaystyle \binom 5 3 = 10$.

In general, $\displaystyle \binom n r = \dfrac{n!}{r!(n-r)!}$, where $r \le n$, $n! = n \times (n-1) \times ... \times 3 \times 2 \times 1$, and $0! = 1$.

It is not until $n = 23$, that a value exceeds one-million: $\displaystyle \binom {23} {10} = 1144066$.

How many, not necessarily distinct, values of $\displaystyle \binom n r$ for $1 \le n \le 100$, are greater than one-million?

Answer: 4075


In the card game poker, a hand consists of five cards and are ranked, from lowest to highest, in the following way:

- High Card: Highest value card.

- One Pair: Two cards of the same value.

- Two Pairs: Two different pairs.

- Three of a Kind: Three cards of the same value.

- Straight: All cards are consecutive values.

- Flush: All cards of the same suit.

- Full House: Three of a kind and a pair.

- Four of a Kind: Four cards of the same value.

- Straight Flush: All cards are consecutive values of same suit.

- Royal Flush: Ten, Jack, Queen, King, Ace, in same suit.

The cards are valued in the order:
2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace.

If two players have the same ranked hands then the rank made up of the highest value wins; for example, a pair of eights beats a pair of fives (see example 1 below). But if two ranks tie, for example, both players have a pair of queens, then highest cards in each hand are compared (see example 4 below); if the highest cards tie then the next highest cards are compared, and so on.

Consider the following five hands dealt to two players:

| Hand |  | Player 1 |  | Player 2 |  | Winner |
| 1 |  | 5H 5C 6S 7S KDPair of Fives |  | 2C 3S 8S 8D TDPair of Eights |  | Player 2 |
| 2 |  | 5D 8C 9S JS ACHighest card Ace |  | 2C 5C 7D 8S QHHighest card Queen |  | Player 1 |
| 3 |  | 2D 9C AS AH ACThree Aces |  | 3D 6D 7D TD QDFlush  with Diamonds |  | Player 2 |
| 4 |  | 4D 6S 9H QH QCPair of QueensHighest card Nine |  | 3D 6D 7H QD QSPair of QueensHighest card Seven |  | Player 1 |
| 5 |  | 2H 2D 4C 4D 4SFull HouseWith Three Fours |  | 3C 3D 3S 9S 9DFull Housewith Three Threes |  | Player 1 |

The file, poker.txt, contains one-thousand random hands dealt to two players. Each line of the file contains ten cards (separated by a single space): the first five are Player 1's cards and the last five are Player 2's cards. You can assume that all hands are valid (no invalid characters or repeated cards), each player's hand is in no specific order, and in each hand there is a clear winner.

How many hands does Player 1 win?

Answer: 376


If we take $47$, reverse and add, $47 + 74 = 121$, which is palindromic.

Not all numbers produce palindromes so quickly. For example,

$$\begin{align}
349 + 943 &= 1292\\
1292 + 2921 &= 4213\\
4213 + 3124 &= 7337
\end{align}$$
That is, $349$ took three iterations to arrive at a palindrome.

Although no one has proved it yet, it is thought that some numbers, like $196$, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, $10677$ is the first number to be shown to require over fifty iterations before producing a palindrome: $4668731596684224866951378664$ ($53$ iterations, $28$-digits).

Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is $4994$.

How many Lychrel numbers are there below ten-thousand?

NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.

Answer: 249


A googol ($10^{100}$) is a massive number: one followed by one-hundred zeros; $100^{100}$ is almost unimaginably large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only $1$.

Considering natural numbers of the form, $a^b$, where $a, b \lt 100$, what is the maximum digital sum?

Answer: 972


It is possible to show that the square root of two can be expressed as an infinite continued fraction.

$\sqrt 2 =1+ \frac 1 {2+ \frac 1 {2 +\frac 1 {2+ \dots}}}$

By expanding this for the first four iterations, we get:

$1 + \frac 1 2 = \frac  32 = 1.5$

$1 + \frac 1 {2 + \frac 1 2} = \frac 7 5 = 1.4$

$1 + \frac 1 {2 + \frac 1 {2+\frac 1 2}} = \frac {17}{12} = 1.41666 \dots$

$1 + \frac 1 {2 + \frac 1 {2+\frac 1 {2+\frac 1 2}}} = \frac {41}{29} = 1.41379 \dots$

The next three expansions are $\frac {99}{70}$, $\frac {239}{169}$, and $\frac {577}{408}$, but the eighth expansion, $\frac {1393}{985}$, is the first example where the number of digits in the numerator exceeds the number of digits in the denominator.

In the first one-thousand expansions, how many fractions contain a numerator with more digits than the denominator?

Answer: 153


Starting with $1$ and spiralling anticlockwise in the following way, a square spiral with side length $7$ is formed.

37 36 35 34 33 32 31

38 17 16 15 14 13 30

39 18  5  4  3 12 29

40 19  6  1  2 11 28

41 20  7  8  9 10 27

42 21 22 23 24 25 26
43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that $8$ out of the $13$ numbers lying along both diagonals are prime; that is, a ratio of $8/13 \approx 62\%$.

If one complete new layer is wrapped around the spiral above, a square spiral with side length $9$ will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below $10\%$?

Answer: 26241


Each character on a computer is assigned a unique code and the preferred standard is ASCII (American Standard Code for Information Interchange). For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.

A modern encryption method is to take a text file, convert the bytes to ASCII, then XOR each byte with a given value, taken from a secret key. The advantage with the XOR function is that using the same encryption key on the cipher text, restores the plain text; for example, 65 XOR 42 = 107, then 107 XOR 42 = 65.

For unbreakable encryption, the key is the same length as the plain text message, and the key is made up of random bytes. The user would keep the encrypted message and the encryption key in different locations, and without both "halves", it is impossible to decrypt the message.

Unfortunately, this method is impractical for most users, so the modified method is to use a password as a key. If the password is shorter than the message, which is likely, the key is repeated cyclically throughout the message. The balance for this method is using a sufficiently long password key for security, but short enough to be memorable.

Your task has been made easy, as the encryption key consists of three lower case characters. Using 0059_cipher.txt (right click and 'Save Link/Target As...'), a file containing the encrypted ASCII codes, and the knowledge that the plain text must contain common English words, decrypt the message and find the sum of the ASCII values in the original text.

Answer: 129448


The primes $3$, $7$, $109$, and $673$, are quite remarkable. By taking any two primes and concatenating them in any order the result will always be prime. For example, taking $7$ and $109$, both $7109$ and $1097$ are prime. The sum of these four primes, $792$, represents the lowest sum for a set of four primes with this property.

Find the lowest sum for a set of five primes for which any two primes concatenate to produce another prime.

Answer: 26033


Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

| Triangle |  | $P_{3,n}=n(n+1)/2$ |  | $1, 3, 6, 10, 15, \dots$ |
| Square |  | $P_{4,n}=n^2$ |  | $1, 4, 9, 16, 25, \dots$ |
| Pentagonal |  | $P_{5,n}=n(3n-1)/2$ |  | $1, 5, 12, 22, 35, \dots$ |
| Hexagonal |  | $P_{6,n}=n(2n-1)$ |  | $1, 6, 15, 28, 45, \dots$ |
| Heptagonal |  | $P_{7,n}=n(5n-3)/2$ |  | $1, 7, 18, 34, 55, \dots$ |
| Octagonal |  | $P_{8,n}=n(3n-2)$ |  | $1, 8, 21, 40, 65, \dots$ |

The ordered set of three $4$-digit numbers: $8128$, $2882$, $8281$, has three interesting properties.

- The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).

- Each polygonal type: triangle ($P_{3,127}=8128$), square ($P_{4,91}=8281$), and pentagonal ($P_{5,44}=2882$), is represented by a different number in the set.

- This is the only set of $4$-digit numbers with this property.

Find the sum of the only ordered set of six cyclic $4$-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.

Answer: 28684


The cube, $41063625$ ($345^3$), can be permuted to produce two other cubes: $56623104$ ($384^3$) and $66430125$ ($405^3$). In fact, $41063625$ is the smallest cube which has exactly three permutations of its digits which are also cube.

Find the smallest cube for which exactly five permutations of its digits are cube.

Answer: 127035954683


The $5$-digit number, $16807=7^5$, is also a fifth power. Similarly, the $9$-digit number, $134217728=8^9$, is a ninth power.

How many $n$-digit positive integers exist which are also an $n$th power?

Answer: 49


All square roots are periodic when written as continued fractions and can be written in the form:

$$\sqrt{N}=a_0 + \dfrac 1 {a_1 + \dfrac 1 {a_2 + \dfrac 1 {a_3 + \dots}}}$$

For example, let us consider $\sqrt{23}:$

$$\sqrt{23} = 4 + \sqrt{23}-4=4 + \dfrac 1 {\dfrac 1 {\sqrt{23}-4}} = 4+\dfrac 1  {1 + \dfrac{\sqrt{23}-3}7}$$

If we continue we would get the following expansion:

$$\sqrt{23}=4 + \dfrac 1 {1 + \dfrac 1 {3+ \dfrac 1 {1 + \dfrac 1 {8+ \dots}}}}$$

The process can be summarised as follows:

$$\begin{align} \quad \quad a_0 &= 4, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7 \\
\quad \quad a_1 &= 1, \frac 7 {\sqrt{23}-3}=\frac {7(\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2 \\
\quad \quad a_2 &= 3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7 \\
\quad \quad a_3 &= 1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} 7=8+\sqrt{23}-4 \\
\quad \quad a_4 &= 8, \frac 1 {\sqrt{23}-4}=\frac {\sqrt{23}+4} 7=1+\frac {\sqrt{23}-3} 7 \\
\quad \quad a_5 &= 1, \frac 7 {\sqrt{23}-3}=\frac {7 (\sqrt{23}+3)} {14}=3+\frac {\sqrt{23}-3} 2 \\
\quad \quad a_6 &= 3, \frac 2 {\sqrt{23}-3}=\frac {2(\sqrt{23}+3)} {14}=1+\frac {\sqrt{23}-4} 7 \\
\quad \quad a_7 &= 1, \frac 7 {\sqrt{23}-4}=\frac {7(\sqrt{23}+4)} {7}=8+\sqrt{23}-4 \end{align}$$

It can be seen that the sequence is repeating. For conciseness, we use the notation $\sqrt{23}=[4;(1,3,1,8)]$, to indicate that the block (1,3,1,8) repeats indefinitely.

The first ten continued fraction representations of (irrational) square roots are:

$\quad \quad \sqrt{2}=[1;(2)]$, period=$1$

$\quad \quad \sqrt{3}=[1;(1,2)]$, period=$2$

$\quad \quad \sqrt{5}=[2;(4)]$, period=$1$

$\quad \quad \sqrt{6}=[2;(2,4)]$, period=$2$

$\quad \quad \sqrt{7}=[2;(1,1,1,4)]$, period=$4$

$\quad \quad \sqrt{8}=[2;(1,4)]$, period=$2$

$\quad \quad \sqrt{10}=[3;(6)]$, period=$1$

$\quad \quad \sqrt{11}=[3;(3,6)]$, period=$2$

$\quad \quad \sqrt{12}=[3;(2,6)]$, period=$2$

$\quad \quad \sqrt{13}=[3;(1,1,1,1,6)]$, period=$5$

Exactly four continued fractions, for $N \le 13$, have an odd period.

How many continued fractions for $N \le 10\,000$ have an odd period?

Answer: 1322


The square root of $2$ can be written as an infinite continued fraction.

$$\sqrt{2} = 1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + ...}}}}$$

The infinite continued fraction can be written, $\sqrt{2} = [1; (2)]$, $(2)$ indicates that $2$ repeats ad infinitum. In a similar way, $\sqrt{23} = [4; (1, 3, 1, 8)]$.

It turns out that the sequence of partial values of continued fractions for square roots provide the best rational approximations. Let us consider the convergents for $\sqrt{2}$.

$$\begin{align}
&1 + \dfrac{1}{2} &= \dfrac{3}{2} \\
&1 + \dfrac{1}{2 + \dfrac{1}{2}} &= \dfrac{7}{5}\\
&1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}} &= \dfrac{17}{12}\\
&1 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2 + \dfrac{1}{2}}}} &= \dfrac{41}{29}
\end{align}$$

Hence the sequence of the first ten convergents for $\sqrt{2}$ are:

$$1, \dfrac{3}{2}, \dfrac{7}{5}, \dfrac{17}{12}, \dfrac{41}{29}, \dfrac{99}{70}, \dfrac{239}{169}, \dfrac{577}{408}, \dfrac{1393}{985}, \dfrac{3363}{2378}, ...$$

What is most surprising is that the important mathematical constant,

$$e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, ... , 1, 2k, 1, ...]$$

The first ten terms in the sequence of convergents for $e$ are:

$$2, 3, \dfrac{8}{3}, \dfrac{11}{4}, \dfrac{19}{7}, \dfrac{87}{32}, \dfrac{106}{39}, \dfrac{193}{71}, \dfrac{1264}{465}, \dfrac{1457}{536}, ...$$

The sum of digits in the numerator of the $10$th convergent is $1 + 4 + 5 + 7 = 17$.

Find the sum of digits in the numerator of the $100$th convergent of the continued fraction for $e$.

Answer: 272


Consider quadratic Diophantine equations of the form:
$$x^2 - Dy^2 = 1$$

For example, when $D=13$, the minimal solution in $x$ is $649^2 - 13 \times 180^2 = 1$.

It can be assumed that there are no solutions in positive integers when $D$ is square.

By finding minimal solutions in $x$ for $D = \{2, 3, 5, 6, 7\}$, we obtain the following:

$$\begin{align}
3^2 - 2 \times 2^2 &= 1\\
2^2 - 3 \times 1^2 &= 1\\
{\color{red}{\mathbf 9}}^2 - 5 \times 4^2 &= 1\\
5^2 - 6 \times 2^2 &= 1\\
8^2 - 7 \times 3^2 &= 1
\end{align}$$
Hence, by considering minimal solutions in $x$ for $D \le 7$, the largest $x$ is obtained when $D=5$.

Find the value of $D \le 1000$ in minimal solutions of $x$ for which the largest value of $x$ is obtained.

Answer: 661


By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4

2 4 6

8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.

NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are $2^{99}$ altogether! If you could check one trillion ($10^{12}$) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;o)

Answer: 7273


Consider the following "magic" 3-gon ring, filled with the numbers 1 to 6, and each line adding to nine.

Working clockwise, and starting from the group of three with the numerically lowest external node (4,3,2 in this example), each solution can be described uniquely. For example, the above solution can be described by the set: 4,3,2; 6,2,1; 5,1,3.

It is possible to complete the ring with four different totals: 9, 10, 11, and 12. There are eight solutions in total.

| Total | Solution Set |
| 9 | 4,2,3; 5,3,1; 6,1,2 |
| 9 | 4,3,2; 6,2,1; 5,1,3 |
| 10 | 2,3,5; 4,5,1; 6,1,3 |
| 10 | 2,5,3; 6,3,1; 4,1,5 |
| 11 | 1,4,6; 3,6,2; 5,2,4 |
| 11 | 1,6,4; 5,4,2; 3,2,6 |
| 12 | 1,5,6; 2,6,4; 3,4,5 |
| 12 | 1,6,5; 3,5,4; 2,4,6 |

By concatenating each group it is possible to form 9-digit strings; the maximum string for a 3-gon ring is 432621513.

Using the numbers 1 to 10, and depending on arrangements, it is possible to form 16- and 17-digit strings. What is the maximum 16-digit string for a "magic" 5-gon ring?

Answer: 6531031914842725


Euler's totient function, $\phi(n)$ [sometimes called the phi function], is defined as the number of positive integers not exceeding $n$ which are relatively prime to $n$. For example, as $1$, $2$, $4$, $5$, $7$, and $8$, are all less than or equal to nine and relatively prime to nine, $\phi(9)=6$.

| $n$ | Relatively Prime | $\phi(n)$ | $n/\phi(n)$ |
| 2 | 1 | 1 | 2 |
| 3 | 1,2 | 2 | 1.5 |
| 4 | 1,3 | 2 | 2 |
| 5 | 1,2,3,4 | 4 | 1.25 |
| 6 | 1,5 | 2 | 3 |
| 7 | 1,2,3,4,5,6 | 6 | 1.1666... |
| 8 | 1,3,5,7 | 4 | 2 |
| 9 | 1,2,4,5,7,8 | 6 | 1.5 |
| 10 | 1,3,7,9 | 4 | 2.5 |

It can be seen that $n = 6$ produces a maximum $n/\phi(n)$ for $n\leq 10$.

Find the value of $n\leq 1\,000\,000$ for which $n/\phi(n)$ is a maximum.

Answer: 510510


Euler's totient function, $\phi(n)$ [sometimes called the phi function], is used to determine the number of positive numbers less than or equal to $n$ which are relatively prime to $n$. For example, as $1, 2, 4, 5, 7$, and $8$, are all less than nine and relatively prime to nine, $\phi(9)=6$.
The number $1$ is considered to be relatively prime to every positive number, so $\phi(1)=1$.

Interestingly, $\phi(87109)=79180$, and it can be seen that $87109$ is a permutation of $79180$.

Find the value of $n$, $1 \lt n \lt 10^7$, for which $\phi(n)$ is a permutation of $n$ and the ratio $n/\phi(n)$ produces a minimum.

Answer: 8319823


Consider the fraction, $\dfrac n d$, where $n$ and $d$ are positive integers. If $n \lt d$ and $\operatorname{HCF}(n,d)=1$, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for $d \le 8$ in ascending order of size, we get:
$$\frac 1 8, \frac 1 7, \frac 1 6, \frac 1 5, \frac 1 4, \frac 2 7, \frac 1 3, \frac 3 8, \mathbf{\frac 2 5}, \frac 3 7, \frac 1 2, \frac 4 7, \frac 3 5, \frac 5 8, \frac 2 3, \frac 5 7, \frac 3 4, \frac 4 5, \frac 5 6, \frac 6 7, \frac 7 8$$

It can be seen that $\dfrac 2 5$ is the fraction immediately to the left of $\dfrac 3 7$.

By listing the set of reduced proper fractions for $d \le 1\,000\,000$ in ascending order of size, find the numerator of the fraction immediately to the left of $\dfrac 3 7$.

Answer: 428570


Consider the fraction, $\dfrac n d$, where $n$ and $d$ are positive integers. If $n \lt d$ and $\operatorname{HCF}(n,d)=1$, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for $d \le 8$ in ascending order of size, we get:
$$\frac 1 8, \frac 1 7, \frac 1 6, \frac 1 5, \frac 1 4, \frac 2 7, \frac 1 3, \frac 3 8, \frac 2 5, \frac 3 7, \frac 1 2, \frac 4 7, \frac 3 5, \frac 5 8, \frac 2 3, \frac 5 7, \frac 3 4, \frac 4 5, \frac 5 6, \frac 6 7, \frac 7 8$$

It can be seen that there are $21$ elements in this set.

How many elements would be contained in the set of reduced proper fractions for $d \le 1\,000\,000$?

Answer: 303963552391


Consider the fraction, $\dfrac n d$, where $n$ and $d$ are positive integers. If $n \lt d$ and $\operatorname{HCF}(n, d)=1$, it is called a reduced proper fraction.

If we list the set of reduced proper fractions for $d \le 8$ in ascending order of size, we get:
$$\frac 1 8, \frac 1 7, \frac 1 6, \frac 1 5, \frac 1 4, \frac 2 7, \frac 1 3, \mathbf{\frac 3 8, \frac 2 5, \frac 3 7}, \frac 1 2, \frac 4 7, \frac 3 5, \frac 5 8, \frac 2 3, \frac 5 7, \frac 3 4, \frac 4 5, \frac 5 6, \frac 6 7, \frac 7 8$$

It can be seen that there are $3$ fractions between $\dfrac 1 3$ and $\dfrac 1 2$.

How many fractions lie between $\dfrac 1 3$ and $\dfrac 1 2$ in the sorted set of reduced proper fractions for $d \le 12\,000$?

Answer: 7295372


The number $145$ is well known for the property that the sum of the factorial of its digits is equal to $145$:
$$1! + 4! + 5! = 1 + 24 + 120 = 145.$$

Perhaps less well known is $169$, in that it produces the longest chain of numbers that link back to $169$; it turns out that there are only three such loops that exist:

$$\begin{align}
&169 \to 363601 \to 1454 \to 169\\
&871 \to 45361 \to 871\\
&872 \to 45362 \to 872
\end{align}$$
It is not difficult to prove that EVERY starting number will eventually get stuck in a loop. For example,

$$\begin{align}
&69 \to 363600 \to 1454 \to 169 \to 363601 (\to 1454)\\
&78 \to 45360 \to 871 \to 45361 (\to 871)\\
&540 \to 145 (\to 145)
\end{align}$$
Starting with $69$ produces a chain of five non-repeating terms, but the longest non-repeating chain with a starting number below one million is sixty terms.

How many chains, with a starting number below one million, contain exactly sixty non-repeating terms?

Answer: 402


It turns out that $\pu{12 cm}$ is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way, but there are many more examples.

- $\pu{\mathbf{12} \mathbf{cm}}$: $(3,4,5)$

- $\pu{\mathbf{24} \mathbf{cm}}$: $(6,8,10)$

- $\pu{\mathbf{30} \mathbf{cm}}$: $(5,12,13)$

- $\pu{\mathbf{36} \mathbf{cm}}$: $(9,12,15)$

- $\pu{\mathbf{40} \mathbf{cm}}$: $(8,15,17)$

- $\pu{\mathbf{48} \mathbf{cm}}$: $(12,16,20)$

In contrast, some lengths of wire, like $\pu{20 cm}$, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using $\pu{120 cm}$ it is possible to form exactly three different integer sided right angle triangles.

- $\pu{\mathbf{120} \mathbf{cm}}$: $(30,40,50)$, $(20,48,52)$, $(24,45,51)$

Given that $L$ is the length of the wire, for how many values of $L \le 1\,500\,000$ can exactly one integer sided right angle triangle be formed?

Answer: 161667


It is possible to write five as a sum in exactly six different ways:

$$\begin{align}
&4 + 1\\
&3 + 2\\
&3 + 1 + 1\\
&2 + 2 + 1\\
&2 + 1 + 1 + 1\\
&1 + 1 + 1 + 1 + 1
\end{align}$$
How many different ways can one hundred be written as a sum of at least two positive integers?

Answer: 190569291


It is possible to write ten as the sum of primes in exactly five different ways:

$$\begin{align}
&7 + 3\\
&5 + 5\\
&5 + 3 + 2\\
&3 + 3 + 2 + 2\\
&2 + 2 + 2 + 2 + 2
\end{align}$$
What is the first value which can be written as the sum of primes in over five thousand different ways?

Answer: 71


Let $p(n)$ represent the number of different ways in which $n$ coins can be separated into piles. For example, five coins can be separated into piles in exactly seven different ways, so $p(5)=7$.

OOOOO

OOOO   O

OOO   OO

OOO   O   O

OO   OO   O

OO   O   O   O

O   O   O   O   O

Find the least value of $n$ for which $p(n)$ is divisible by one million.

Answer: 55374


A common security method used for online banking is to ask the user for three random characters from a passcode. For example, if the passcode was 531278, they may ask for the 2nd, 3rd, and 5th characters; the expected reply would be: 317.

The text file, keylog.txt, contains fifty successful login attempts.

Given that the three characters are always asked for in order, analyse the file so as to determine the shortest possible secret passcode of unknown length.

Answer: 73162890


It is well known that if the square root of a natural number is not an integer, then it is irrational. The decimal expansion of such square roots is infinite without any repeating pattern at all.

The square root of two is $1.41421356237309504880\cdots$, and the digital sum of the first one hundred decimal digits is $475$.

For the first one hundred natural numbers, find the total of the digital sums of the first one hundred decimal digits for all the irrational square roots.

Answer: 40886


In the $5$ by $5$ matrix below, the minimal path sum from the top left to the bottom right, by only moving to the right and down, is indicated in bold red and is equal to $2427$.

$$
\begin{pmatrix}
\color{red}{131} & 673 & 234 & 103 & 18\\
\color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\
630 & 803 & \color{red}{746} & \color{red}{422} & 111\\
537 & 699 & 497 & \color{red}{121} & 956\\
805 & 732 & 524 & \color{red}{37} & \color{red}{331}
\end{pmatrix}
$$

Find the minimal path sum from the top left to the bottom right by only moving right and down in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing an $80$ by $80$ matrix.

Answer: 427337


NOTE: This problem is a more challenging version of Problem 81.

The minimal path sum in the $5$ by $5$ matrix below, by starting in any cell in the left column and finishing in any cell in the right column, and only moving up, down, and right, is indicated in red and bold; the sum is equal to $994$.

$$
\begin{pmatrix}
131 & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\
\color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & 150\\
630 & 803 & 746 & 422 & 111\\
537 & 699 & 497 & 121 & 956\\
805 & 732 & 524 & 37 & 331
\end{pmatrix}
$$

Find the minimal path sum from the left column to the right column in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing an $80$ by $80$ matrix.

Answer: 260324


NOTE: This problem is a significantly more challenging version of Problem 81.

In the $5$ by $5$ matrix below, the minimal path sum from the top left to the bottom right, by moving left, right, up, and down, is indicated in bold red and is equal to $2297$.

$$
\begin{pmatrix}
\color{red}{131} & 673 & \color{red}{234} & \color{red}{103} & \color{red}{18}\\
\color{red}{201} & \color{red}{96} & \color{red}{342} & 965 & \color{red}{150}\\
630 & 803 & 746 & \color{red}{422} & \color{red}{111}\\
537 & 699 & 497 & \color{red}{121} & 956\\
805 & 732 & 524 & \color{red}{37} & \color{red}{331}
\end{pmatrix}
$$

Find the minimal path sum from the top left to the bottom right by moving left, right, up, and down in matrix.txt (right click and "Save Link/Target As..."), a 31K text file containing an $80$ by $80$ matrix.

Answer: 425185


In the game, Monopoly, the standard board is set up in the following way:

A player starts on the GO square and adds the scores on two 6-sided dice to determine the number of squares they advance in a clockwise direction. Without any further rules we would expect to visit each square with equal probability: 2.5%. However, landing on G2J (Go To Jail), CC (community chest), and CH (chance) changes this distribution.

In addition to G2J, and one card from each of CC and CH, that orders the player to go directly to jail, if a player rolls three consecutive doubles, they do not advance the result of their 3rd roll. Instead they proceed directly to jail.

At the beginning of the game, the CC and CH cards are shuffled. When a player lands on CC or CH they take a card from the top of the respective pile and, after following the instructions, it is returned to the bottom of the pile. There are sixteen cards in each pile, but for the purpose of this problem we are only concerned with cards that order a movement; any instruction not concerned with movement will be ignored and the player will remain on the CC/CH square.

- Community Chest (2/16 cards):

- Advance to GO

- Go to JAIL

- Chance (10/16 cards):

- Advance to GO

- Go to JAIL

- Go to C1

- Go to E3

- Go to H2

- Go to R1

- Go to next R (railway company)

- Go to next R

- Go to next U (utility company)

- Go back 3 squares.

The heart of this problem concerns the likelihood of visiting a particular square. That is, the probability of finishing at that square after a roll. For this reason it should be clear that, with the exception of G2J for which the probability of finishing on it is zero, the CH squares will have the lowest probabilities, as 5/8 request a movement to another square, and it is the final square that the player finishes at on each roll that we are interested in. We shall make no distinction between "Just Visiting" and being sent to JAIL, and we shall also ignore the rule about requiring a double to "get out of jail", assuming that they pay to get out on their next turn.

By starting at GO and numbering the squares sequentially from 00 to 39 we can concatenate these two-digit numbers to produce strings that correspond with sets of squares.

Statistically it can be shown that the three most popular squares, in order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO (3.09%) = Square 00. So these three most popular squares can be listed with the six-digit modal string: 102400.

If, instead of using two 6-sided dice, two 4-sided dice are used, find the six-digit modal string.

Answer: 101524


By counting carefully it can be seen that a rectangular grid measuring $3$ by $2$ contains eighteen rectangles:

Although there exists no rectangular grid that contains exactly two million rectangles, find the area of the grid with the nearest solution.

Answer: 2772


A spider, S, sits in one corner of a cuboid room, measuring $6$ by $5$ by $3$, and a fly, F, sits in the opposite corner. By travelling on the surfaces of the room the shortest "straight line" distance from S to F is $10$ and the path is shown on the diagram.

However, there are up to three "shortest" path candidates for any given cuboid and the shortest route doesn't always have integer length.

It can be shown that there are exactly $2060$ distinct cuboids, ignoring rotations, with integer dimensions, up to a maximum size of $M$ by $M$ by $M$, for which the shortest route has integer length when $M = 100$. This is the least value of $M$ for which the number of solutions first exceeds two thousand; the number of solutions when $M = 99$ is $1975$.

Find the least value of $M$ such that the number of solutions first exceeds one million.

Answer: 1818


The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is $28$. In fact, there are exactly four numbers below fifty that can be expressed in such a way:

$$\begin{align}
28 &= 2^2 + 2^3 + 2^4\\
33 &= 3^2 + 2^3 + 2^4\\
49 &= 5^2 + 2^3 + 2^4\\
47 &= 2^2 + 3^3 + 2^4
\end{align}$$
How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power?

Answer: 1097343


A natural number, $N$, that can be written as the sum and product of a given set of at least two natural numbers, $\{a_1, a_2, \dots, a_k\}$ is called a product-sum number: $N = a_1 + a_2 + \cdots + a_k = a_1 \times a_2 \times \cdots \times a_k$.

For example, $6 = 1 + 2 + 3 = 1 \times 2 \times 3$.

For a given set of size, $k$, we shall call the smallest $N$ with this property a minimal product-sum number. The minimal product-sum numbers for sets of size, $k = 2, 3, 4, 5$, and $6$ are as follows.

- $k=2$: $4 = 2 \times 2 = 2 + 2$

- $k=3$: $6 = 1 \times 2 \times 3 = 1 + 2 + 3$

- $k=4$: $8 = 1 \times 1 \times 2 \times 4 = 1 + 1 + 2 + 4$

- $k=5$: $8 = 1 \times 1 \times 2 \times 2 \times 2 = 1 + 1 + 2 + 2 + 2$
- $k=6$: $12 = 1 \times 1 \times 1 \times 1 \times 2 \times 6 = 1 + 1 + 1 + 1 + 2 + 6$

Hence for $2 \le k \le 6$, the sum of all the minimal product-sum numbers is $4+6+8+12 = 30$; note that $8$ is only counted once in the sum.

In fact, as the complete set of minimal product-sum numbers for $2 \le k \le 12$ is $\{4, 6, 8, 12, 15, 16\}$, the sum is $61$.

What is the sum of all the minimal product-sum numbers for $2 \le k \le 12000$?

Answer: 7587457


For a number written in Roman numerals to be considered valid there are basic rules which must be followed. Even though the rules allow some numbers to be expressed in more than one way there is always a "best" way of writing a particular number.

For example, it would appear that there are at least six ways of writing the number sixteen:

IIIIIIIIIIIIIIII

VIIIIIIIIIII

VVIIIIII

XIIIIII

VVVI

XVI

However, according to the rules only XIIIIII and XVI are valid, and the last example is considered to be the most efficient, as it uses the least number of numerals.

The 11K text file, roman.txt (right click and 'Save Link/Target As...'), contains one thousand numbers written in valid, but not necessarily minimal, Roman numerals; see About... Roman Numerals for the definitive rules for this problem.

Find the number of characters saved by writing each of these in their minimal form.

Note: You can assume that all the Roman numerals in the file contain no more than four consecutive identical units.

Answer: 743


Each of the six faces on a cube has a different digit ($0$ to $9$) written on it; the same is done to a second cube. By placing the two cubes side-by-side in different positions we can form a variety of $2$-digit numbers.

For example, the square number $64$ could be formed:

In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred: $01$, $04$, $09$, $16$, $25$, $36$, $49$, $64$, and $81$.

For example, one way this can be achieved is by placing $\{0, 5, 6, 7, 8, 9\}$ on one cube and $\{1, 2, 3, 4, 8, 9\}$ on the other cube.

However, for this problem we shall allow the $6$ or $9$ to be turned upside-down so that an arrangement like $\{0, 5, 6, 7, 8, 9\}$ and $\{1, 2, 3, 4, 6, 7\}$ allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain $09$.

In determining a distinct arrangement we are interested in the digits on each cube, not the order.

- $\{1, 2, 3, 4, 5, 6\}$ is equivalent to $\{3, 6, 4, 1, 2, 5\}$

- $\{1, 2, 3, 4, 5, 6\}$ is distinct from $\{1, 2, 3, 4, 5, 9\}$

But because we are allowing $6$ and $9$ to be reversed, the two distinct sets in the last example both represent the extended set $\{1, 2, 3, 4, 5, 6, 9\}$ for the purpose of forming $2$-digit numbers.

How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?

Answer: 1217


The points $P(x_1, y_1)$ and $Q(x_2, y_2)$ are plotted at integer co-ordinates and are joined to the origin, $O(0,0)$, to form $\triangle OPQ$.

There are exactly fourteen triangles containing a right angle that can be formed when each co-ordinate lies between $0$ and $2$ inclusive; that is, $0 \le x_1, y_1, x_2, y_2 \le 2$.

Given that $0 \le x_1, y_1, x_2, y_2 \le 50$, how many right triangles can be formed?

Answer: 14234


A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.

For example,
$$\begin{align}
&44 \to 32 \to 13 \to 10 \to \mathbf 1 \to \mathbf 1\\
&85 \to \mathbf{89} \to 145 \to 42 \to 20 \to 4 \to 16 \to 37 \to 58 \to \mathbf{89}
\end{align}$$

Therefore any chain that arrives at $1$ or $89$ will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at $1$ or $89$.

How many starting numbers below ten million will arrive at $89$?

Answer: 8581146


By using each of the digits from the set, $\{1, 2, 3, 4\}$, exactly once, and making use of the four arithmetic operations ($+, -, \times, /$) and brackets/parentheses, it is possible to form different positive integer targets.

For example,

$$\begin{align}
8 &= (4 \times (1 + 3)) / 2\\
14 &= 4 \times (3 + 1 / 2)\\
19 &= 4 \times (2 + 3) - 1\\
36 &= 3 \times 4 \times (2 + 1)
\end{align}$$
Note that concatenations of the digits, like $12 + 34$, are not allowed.

Using the set, $\{1, 2, 3, 4\}$, it is possible to obtain thirty-one different target numbers of which $36$ is the maximum, and each of the numbers $1$ to $28$ can be obtained before encountering the first non-expressible number.

Find the set of four distinct digits, $a \lt b \lt c \lt d$, for which the longest set of consecutive positive integers, $1$ to $n$, can be obtained, giving your answer as a string: abcd.

Answer: 1258


It is easily proved that no equilateral triangle exists with integral length sides and integral area. However, the almost equilateral triangle $5$-$5$-$6$ has an area of $12$ square units.

We shall define an almost equilateral triangle to be a triangle for which two sides are equal and the third differs by no more than one unit.

Find the sum of the perimeters of all almost equilateral triangles with integral side lengths and area and whose perimeters do not exceed one billion ($1\,000\,000\,000$).

Answer: 518408346


The proper divisors of a number are all the divisors excluding the number itself. For example, the proper divisors of $28$ are $1$, $2$, $4$, $7$, and $14$. As the sum of these divisors is equal to $28$, we call it a perfect number.

Interestingly the sum of the proper divisors of $220$ is $284$ and the sum of the proper divisors of $284$ is $220$, forming a chain of two numbers. For this reason, $220$ and $284$ are called an amicable pair.

Perhaps less well known are longer chains. For example, starting with $12496$, we form a chain of five numbers:
$$12496 \to 14288 \to 15472 \to 14536 \to 14264 (\to 12496 \to \cdots)$$

Since this chain returns to its starting point, it is called an amicable chain.

Find the smallest member of the longest amicable chain with no element exceeding one million.

Answer: 14316


Su Doku (Japanese meaning number place) is the name given to a popular puzzle concept. Its origin is unclear, but credit must be attributed to Leonhard Euler who invented a similar, and much more difficult, puzzle idea called Latin Squares. The objective of Su Doku puzzles, however, is to replace the blanks (or zeros) in a 9 by 9 grid in such that each row, column, and 3 by 3 box contains each of the digits 1 to 9. Below is an example of a typical starting puzzle grid and its solution grid.

A well constructed Su Doku puzzle has a unique solution and can be solved by logic, although it may be necessary to employ "guess and test" methods in order to eliminate options (there is much contested opinion over this). The complexity of the search determines the difficulty of the puzzle; the example above is considered easy because it can be solved by straight forward direct deduction.

The 6K text file, sudoku.txt (right click and 'Save Link/Target As...'), contains fifty different Su Doku puzzles ranging in difficulty, but all with unique solutions (the first puzzle in the file is the example above).

By solving all fifty puzzles find the sum of the 3-digit numbers found in the top left corner of each solution grid; for example, 483 is the 3-digit number found in the top left corner of the solution grid above.

Answer: 24702


The first known prime found to exceed one million digits was discovered in 1999, and is a Mersenne prime of the form $2^{6972593} - 1$; it contains exactly $2\,098\,960$ digits. Subsequently other Mersenne primes, of the form $2^p - 1$, have been found which contain more digits.

However, in 2004 there was found a massive non-Mersenne prime which contains $2\,357\,207$ digits: $28433 \times 2^{7830457} + 1$.

Find the last ten digits of this prime number.

Answer: 8739992577


By replacing each of the letters in the word CARE with $1$, $2$, $9$, and $6$ respectively, we form a square number: $1296 = 36^2$. What is remarkable is that, by using the same digital substitutions, the anagram, RACE, also forms a square number: $9216 = 96^2$. We shall call CARE (and RACE) a square anagram word pair and specify further that leading zeroes are not permitted, neither may a different letter have the same digital value as another letter.

Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words, find all the square anagram word pairs (a palindromic word is NOT considered to be an anagram of itself).

What is the largest square number formed by any member of such a pair?

NOTE: All anagrams formed must be contained in the given text file.

Answer: 18769


Comparing two numbers written in index form like $2^{11}$ and $3^7$ is not difficult, as any calculator would confirm that $2^{11} = 2048 \lt 3^7 = 2187$.

However, confirming that $632382^{518061} \gt 519432^{525806}$ would be much more difficult, as both numbers contain over three million digits.

Using base_exp.txt (right click and 'Save Link/Target As...'), a 22K text file containing one thousand lines with a base/exponent pair on each line, determine which line number has the greatest numerical value.

NOTE: The first two lines in the file represent the numbers in the example given above.

Answer: 709


If a box contains twenty-one coloured discs, composed of fifteen blue discs and six red discs, and two discs were taken at random, it can be seen that the probability of taking two blue discs, $P(\text{BB}) = (15/21) \times (14/20) = 1/2$.

The next such arrangement, for which there is exactly $50\%$ chance of taking two blue discs at random, is a box containing eighty-five blue discs and thirty-five red discs.

By finding the first arrangement to contain over $10^{12} = 1\,000\,000\,000\,000$ discs in total, determine the number of blue discs that the box would contain.

Answer: 756872327473


If we are presented with the first $k$ terms of a sequence it is impossible to say with certainty the value of the next term, as there are infinitely many polynomial functions that can model the sequence.

As an example, let us consider the sequence of cube numbers. This is defined by the generating function,
$u_n = n^3$: $1, 8, 27, 64, 125, 216, \dots$

Suppose we were only given the first two terms of this sequence. Working on the principle that "simple is best" we should assume a linear relationship and predict the next term to be $15$ (common difference $7$). Even if we were presented with the first three terms, by the same principle of simplicity, a quadratic relationship should be assumed.

We shall define $\operatorname{OP}(k, n)$ to be the $n$th term of the optimum polynomial generating function for the first $k$ terms of a sequence. It should be clear that $\operatorname{OP}(k, n)$ will accurately generate the terms of the sequence for $n \le k$, and potentially the first incorrect term (FIT) will be $\operatorname{OP}(k, k+1)$; in which case we shall call it a bad OP (BOP).

As a basis, if we were only given the first term of sequence, it would be most sensible to assume constancy; that is, for $n \ge 2$, $\operatorname{OP}(1, n) = u_1$.

Hence we obtain the following $\operatorname{OP}$s for the cubic sequence:

| $\operatorname{OP}(1, n) = 1$ | $1, {\color{red}\mathbf 1}, 1, 1, \dots$ |
| $\operatorname{OP}(2, n) = 7n - 6$ | $1, 8, {\color{red}\mathbf{15}}, \dots$ |
| $\operatorname{OP}(3, n) = 6n^2 - 11n + 6$ | $1, 8, 27, {\color{red}\mathbf{58}}, \dots$ |
| $\operatorname{OP}(4, n) = n^3$ | $1, 8, 27, 64, 125, \dots$ |

Clearly no BOPs exist for $k \ge 4$.

By considering the sum of FITs generated by the BOPs (indicated in red above), we obtain $1 + 15 + 58 = 74$.

Consider the following tenth degree polynomial generating function:
$$u_n = 1 - n + n^2 - n^3 + n^4 - n^5 + n^6 - n^7 + n^8 - n^9 + n^{10}.$$

Find the sum of FITs for the BOPs.

Answer: 37076114526


Three distinct points are plotted at random on a Cartesian plane, for which $-1000 \le x, y \le 1000$, such that a triangle is formed.

Consider the following two triangles:

$$\begin{gather}
A(-340,495), B(-153,-910), C(835,-947)\\
X(-175,41), Y(-421,-714), Z(574,-645)
\end{gather}$$
It can be verified that triangle $ABC$ contains the origin, whereas triangle $XYZ$ does not.

Using triangles.txt (right click and 'Save Link/Target As...'), a 27K text file containing the co-ordinates of one thousand "random" triangles, find the number of triangles for which the interior contains the origin.

NOTE: The first two examples in the file represent the triangles in the example given above.

Answer: 228


Let $S(A)$ represent the sum of elements in set $A$ of size $n$. We shall call it a special sum set if for any two non-empty disjoint subsets, $B$ and $C$, the following properties are true:

- $S(B) \ne S(C)$; that is, sums of subsets cannot be equal.

- If $B$ contains more elements than $C$ then $S(B) \gt S(C)$.

If $S(A)$ is minimised for a given $n$, we shall call it an optimum special sum set. The first five optimum special sum sets are given below.

- $n = 1$: $\{1\}$

- $n = 2$: $\{1, 2\}$

- $n = 3$: $\{2, 3, 4\}$

- $n = 4$: $\{3, 5, 6, 7\}$

- $n = 5$: $\{6, 9, 11, 12, 13\}$

It seems that for a given optimum set, $A = \{a_1, a_2, \dots, a_n\}$, the next optimum set is of the form $B = \{b, a_1 + b, a_2 + b, \dots, a_n + b\}$, where $b$ is the "middle" element on the previous row.

By applying this "rule" we would expect the optimum set for $n = 6$ to be $A = \{11, 17, 20, 22, 23, 24\}$, with $S(A) = 117$. However, this is not the optimum set, as we have merely applied an algorithm to provide a near optimum set. The optimum set for $n = 6$ is $A = \{11, 18, 19, 20, 22, 25\}$, with $S(A) = 115$ and corresponding set string: 111819202225.

Given that $A$ is an optimum special sum set for $n = 7$, find its set string.

NOTE: This problem is related to Problem 105 and Problem 106.

Answer: 20313839404245


The Fibonacci sequence is defined by the recurrence relation:

$F_n = F_{n - 1} + F_{n - 2}$, where $F_1 = 1$ and $F_2 = 1$.
It turns out that $F_{541}$, which contains $113$ digits, is the first Fibonacci number for which the last nine digits are $1$-$9$ pandigital (contain all the digits $1$ to $9$, but not necessarily in order). And $F_{2749}$, which contains $575$ digits, is the first Fibonacci number for which the first nine digits are $1$-$9$ pandigital.

Given that $F_k$ is the first Fibonacci number for which the first nine digits AND the last nine digits are $1$-$9$ pandigital, find $k$.

Answer: 329468


Let $S(A)$ represent the sum of elements in set $A$ of size $n$. We shall call it a special sum set if for any two non-empty disjoint subsets, $B$ and $C$, the following properties are true:

- $S(B) \ne S(C)$; that is, sums of subsets cannot be equal.

- If $B$ contains more elements than $C$ then $S(B) \gt S(C)$.

For example, $\{81, 88, 75, 42, 87, 84, 86, 65\}$ is not a special sum set because $65 + 87 + 88 = 75 + 81 + 84$, whereas $\{157, 150, 164, 119, 79, 159, 161, 139, 158\}$ satisfies both rules for all possible subset pair combinations and $S(A) = 1286$.

Using sets.txt (right click and "Save Link/Target As..."), a 4K text file with one-hundred sets containing seven to twelve elements (the two examples given above are the first two sets in the file), identify all the special sum sets, $A_1, A_2, \dots, A_k$, and find the value of $S(A_1) + S(A_2) + \cdots + S(A_k)$.

NOTE: This problem is related to Problem 103 and Problem 106.

Answer: 73702


Let $S(A)$ represent the sum of elements in set $A$ of size $n$. We shall call it a special sum set if for any two non-empty disjoint subsets, $B$ and $C$, the following properties are true:

- $S(B) \ne S(C)$; that is, sums of subsets cannot be equal.

- If $B$ contains more elements than $C$ then $S(B) \gt S(C)$.

For this problem we shall assume that a given set contains $n$ strictly increasing elements and it already satisfies the second rule.

Surprisingly, out of the $25$ possible subset pairs that can be obtained from a set for which $n = 4$, only $1$ of these pairs need to be tested for equality (first rule). Similarly, when $n = 7$, only $70$ out of the $966$ subset pairs need to be tested.

For $n = 12$, how many of the $261625$ subset pairs that can be obtained need to be tested for equality?

NOTE: This problem is related to Problem 103 and Problem 105.

Answer: 21384


The following undirected network consists of seven vertices and twelve edges with a total weight of 243.

The same network can be represented by the matrix below.

|  | A | B | C | D | E | F | G |
| A | - | 16 | 12 | 21 | - | - | - |
| B | 16 | - | - | 17 | 20 | - | - |
| C | 12 | - | - | 28 | - | 31 | - |
| D | 21 | 17 | 28 | - | 18 | 19 | 23 |
| E | - | 20 | - | 18 | - | - | 11 |
| F | - | - | 31 | 19 | - | - | 27 |
| G | - | - | - | 23 | 11 | 27 | - |

However, it is possible to optimise the network by removing some edges and still ensure that all points on the network remain connected. The network which achieves the maximum saving is shown below. It has a weight of 93, representing a saving of 243 − 93 = 150 from the original network.

Using network.txt (right click and 'Save Link/Target As...'), a 6K text file containing a network with forty vertices, and given in matrix form, find the maximum saving which can be achieved by removing redundant edges whilst ensuring that the network remains connected.

Answer: 259679


In the following equation $x$, $y$, and $n$ are positive integers.

$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$$
For $n = 4$ there are exactly three distinct solutions:

$$\begin{align}
\dfrac{1}{5} + \dfrac{1}{20} &= \dfrac{1}{4}\\
\dfrac{1}{6} + \dfrac{1}{12} &= \dfrac{1}{4}\\
\dfrac{1}{8} + \dfrac{1}{8} &= \dfrac{1}{4}
\end{align}
$$

What is the least value of $n$ for which the number of distinct solutions exceeds one-thousand?

NOTE: This problem is an easier version of Problem 110; it is strongly advised that you solve this one first.

Answer: 180180


In the game of darts a player throws three darts at a target board which is split into twenty equal sized sections numbered one to twenty.

The score of a dart is determined by the number of the region that the dart lands in. A dart landing outside the red/green outer ring scores zero. The black and cream regions inside this ring represent single scores. However, the red/green outer ring and middle ring score double and treble scores respectively.

At the centre of the board are two concentric circles called the bull region, or bulls-eye. The outer bull is worth 25 points and the inner bull is a double, worth 50 points.

There are many variations of rules but in the most popular game the players will begin with a score 301 or 501 and the first player to reduce their running total to zero is a winner. However, it is normal to play a "doubles out" system, which means that the player must land a double (including the double bulls-eye at the centre of the board) on their final dart to win; any other dart that would reduce their running total to one or lower means the score for that set of three darts is "bust".

When a player is able to finish  on their current score it is called a "checkout" and the highest checkout is 170: T20 T20 D25 (two treble 20s and double bull).

There are exactly eleven distinct ways to checkout on a score of 6:

|  |  |  |
| D3 |  |  |
| D1 | D2 |  |
| S2 | D2 |  |
| D2 | D1 |  |
| S4 | D1 |  |
| S1 | S1 | D2 |
| S1 | T1 | D1 |
| S1 | S3 | D1 |
| D1 | D1 | D1 |
| D1 | S2 | D1 |
| S2 | S2 | D1 |

Note that D1 D2 is considered different to D2 D1 as they finish on different doubles. However, the combination S1 T1 D1 is considered the same as T1 S1 D1.

In addition we shall not include misses in considering combinations; for example, D3 is the same as 0 D3 and 0 0 D3.

Incredibly there are 42336 distinct ways of checking out in total.

How many distinct ways can a player checkout with a score less than 100?

Answer: 38182


In the following equation $x$, $y$, and $n$ are positive integers.

$$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$$

It can be verified that when $n = 1260$ there are $113$ distinct solutions and this is the least value of $n$ for which the total number of distinct solutions exceeds one hundred.

What is the least value of $n$ for which the number of distinct solutions exceeds four million?

NOTE: This problem is a much more difficult version of Problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.

Answer: 9350130049860600


Considering $4$-digit primes containing repeated digits it is clear that they cannot all be the same: $1111$ is divisible by $11$, $2222$ is divisible by $22$, and so on. But there are nine $4$-digit primes containing three ones:
$$1117, 1151, 1171, 1181, 1511, 1811, 2111, 4111, 8111.$$

We shall say that $M(n, d)$ represents the maximum number of repeated digits for an $n$-digit prime where $d$ is the repeated digit, $N(n, d)$ represents the number of such primes, and $S(n, d)$ represents the sum of these primes.

So $M(4, 1) = 3$ is the maximum number of repeated digits for a $4$-digit prime where one is the repeated digit, there are $N(4, 1) = 9$ such primes, and the sum of these primes is $S(4, 1) = 22275$. It turns out that for $d = 0$, it is only possible to have $M(4, 0) = 2$ repeated digits, but there are $N(4, 0) = 13$ such cases.

In the same way we obtain the following results for $4$-digit primes.

| Digit, d | M(4, d) | N(4, d) | S(4, d) |
| 0 | 2 | 13 | 67061 |
| 1 | 3 | 9 | 22275 |
| 2 | 3 | 1 | 2221 |
| 3 | 3 | 12 | 46214 |
| 4 | 3 | 2 | 8888 |
| 5 | 3 | 1 | 5557 |
| 6 | 3 | 1 | 6661 |
| 7 | 3 | 9 | 57863 |
| 8 | 3 | 1 | 8887 |
| 9 | 3 | 7 | 48073 |

For $d = 0$ to $9$, the sum of all $S(4, d)$ is $273700$.

Find the sum of all $S(10, d)$.

Answer: 612407567715


Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, $134468$.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, $66420$.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, $155349$.

Clearly there cannot be any bouncy numbers below one-hundred, but just over half of the numbers below one-thousand ($525$) are bouncy. In fact, the least number for which the proportion of bouncy numbers first reaches $50\%$ is $538$.

Surprisingly, bouncy numbers become more and more common and by the time we reach $21780$ the proportion of bouncy numbers is equal to $90\%$.

Find the least number for which the proportion of bouncy numbers is exactly $99\%$.

Answer: 1587000


Working from left-to-right if no digit is exceeded by the digit to its left it is called an increasing number; for example, $134468$.

Similarly if no digit is exceeded by the digit to its right it is called a decreasing number; for example, $66420$.

We shall call a positive integer that is neither increasing nor decreasing a "bouncy" number; for example, $155349$.

As $n$ increases, the proportion of bouncy numbers below $n$ increases such that there are only $12951$ numbers below one-million that are not bouncy and only $277032$ non-bouncy numbers below $10^{10}$.

How many numbers below a googol ($10^{100}$) are not bouncy?

Answer: 51161058134250


A row measuring seven units in length has red blocks with a minimum length of three units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one grey square. There are exactly seventeen ways of doing this.

How many ways can a row measuring fifty units in length be filled?

NOTE: Although the example above does not lend itself to the possibility, in general it is permitted to mix block sizes. For example, on a row measuring eight units in length you could use red (3), grey (1), and red (4).

Answer: 16475640049


NOTE: This is a more difficult version of Problem 114.

A row measuring $n$ units in length has red blocks with a minimum length of $m$ units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square.

Let the fill-count function, $F(m, n)$, represent the number of ways that a row can be filled.

For example, $F(3, 29) = 673135$ and $F(3, 30) = 1089155$.

That is, for $m = 3$, it can be seen that $n = 30$ is the smallest value for which the fill-count function first exceeds one million.

In the same way, for $m = 10$, it can be verified that $F(10, 56) = 880711$ and $F(10, 57) = 1148904$, so $n = 57$ is the least value for which the fill-count function first exceeds one million.

For $m = 50$, find the least value of $n$ for which the fill-count function first exceeds one million.

Answer: 168


A row of five grey square tiles is to have a number of its tiles replaced with coloured oblong tiles chosen from red (length two), green (length three), or blue (length four).

If red tiles are chosen there are exactly seven ways this can be done.

If green tiles are chosen there are three ways.

And if blue tiles are chosen there are two ways.

Assuming that colours cannot be mixed there are $7 + 3 + 2 = 12$ ways of replacing the grey tiles in a row measuring five units in length.

How many different ways can the grey tiles in a row measuring fifty units in length be replaced if colours cannot be mixed and at least one coloured tile must be used?

NOTE: This is related to Problem 117.

Answer: 20492570929


Using a combination of grey square tiles and oblong tiles chosen from: red tiles (measuring two units), green tiles (measuring three units), and blue tiles (measuring four units), it is possible to tile a row measuring five units in length in exactly fifteen different ways.

How many ways can a row measuring fifty units in length be tiled?

NOTE: This is related to Problem 116.

Answer: 100808458960497


Using all of the digits $1$ through $9$ and concatenating them freely to form decimal integers, different sets can be formed. Interestingly with the set $\{2,5,47,89,631\}$, all of the elements belonging to it are prime.

How many distinct sets containing each of the digits one through nine exactly once contain only prime elements?

Answer: 44680


The number $512$ is interesting because it is equal to the sum of its digits raised to some power: $5 + 1 + 2 = 8$, and $8^3 = 512$. Another example of a number with this property is $614656 = 28^4$.

We shall define $a_n$ to be the $n$th term of this sequence and insist that a number must contain at least two digits to have a sum.

You are given that $a_2 = 512$ and $a_{10} = 614656$.

Find $a_{30}$.

Answer: 248155780267521


Let $r$ be the remainder when $(a - 1)^n + (a + 1)^n$ is divided by $a^2$.

For example, if $a = 7$ and $n = 3$, then $r = 42$: $6^3 + 8^3 = 728 \equiv 42 \mod 49$. And as $n$ varies, so too will $r$, but for $a = 7$ it turns out that $r_{\mathrm{max}} = 42$.

For $3 \le a \le 1000$, find $\sum r_{\mathrm{max}}$.

Answer: 333082500


A bag contains one red disc and one blue disc. In a game of chance a player takes a disc at random and its colour is noted. After each turn the disc is returned to the bag, an extra red disc is added, and another disc is taken at random.

The player pays £1 to play and wins if they have taken more blue discs than red discs at the end of the game.

If the game is played for four turns, the probability of a player winning is exactly 11/120, and so the maximum prize fund the banker should allocate for winning in this game would be £10 before they would expect to incur a loss. Note that any payout will be a whole number of pounds and also includes the original £1 paid to play the game, so in the example given the player actually wins £9.

Find the maximum prize fund that should be allocated to a single game in which fifteen turns are played.

Answer: 2269


The most naive way of computing $n^{15}$ requires fourteen multiplications:
$$n \times n \times \cdots \times n = n^{15}.$$

But using a "binary" method you can compute it in six multiplications:

$$\begin{align}
n \times n &= n^2\\
n^2 \times n^2 &= n^4\\
n^4 \times n^4 &= n^8\\
n^8 \times n^4 &= n^{12}\\
n^{12} \times n^2 &= n^{14}\\
n^{14} \times n &= n^{15}
\end{align}$$
However it is yet possible to compute it in only five multiplications:

$$\begin{align}
n \times n &= n^2\\
n^2 \times n &= n^3\\
n^3 \times n^3 &= n^6\\
n^6 \times n^6 &= n^{12}\\
n^{12} \times n^3 &= n^{15}
\end{align}$$
We shall define $m(k)$ to be the minimum number of multiplications to compute $n^k$; for example $m(15) = 5$.

Find $\sum\limits_{k = 1}^{200} m(k)$.

Answer: 1582


Let $p_n$ be the $n$th prime: $2, 3, 5, 7, 11, \dots$, and let $r$ be the remainder when $(p_n - 1)^n + (p_n + 1)^n$ is divided by $p_n^2$.

For example, when $n = 3$, $p_3 = 5$, and $4^3 + 6^3 = 280 \equiv 5 \mod 25$.

The least value of $n$ for which the remainder first exceeds $10^9$ is $7037$.

Find the least value of $n$ for which the remainder first exceeds $10^{10}$.

Answer: 21035


The radical of $n$, $\operatorname{rad}(n)$, is the product of the distinct prime factors of $n$. For example, $504 = 2^3 \times 3^2 \times 7$, so $\operatorname{rad}(504) = 2 \times 3 \times 7 = 42$.

If we calculate $\operatorname{rad}(n)$ for $1 \le n \le 10$, then sort them on $\operatorname{rad}(n)$, and sorting on $n$ if the radical values are equal, we get:

| Unsorted |  | Sorted |
| --- | --- | --- |
| n | rad(n) |  | n | rad(n) | k |
| 1 | 1 |  | 1 | 1 | 1 |
| 2 | 2 |  | 2 | 2 | 2 |
| 3 | 3 |  | 4 | 2 | 3 |
| 4 | 2 |  | 8 | 2 | 4 |
| 5 | 5 |  | 3 | 3 | 5 |
| 6 | 6 |  | 9 | 3 | 6 |
| 7 | 7 |  | 5 | 5 | 7 |
| 8 | 2 |  | 6 | 6 | 8 |
| 9 | 3 |  | 7 | 7 | 9 |
| 10 | 10 |  | 10 | 10 | 10 |

Let $E(k)$ be the $k$-th element in the sorted $n$ column; for example, $E(4) = 8$ and $E(6) = 9$.

If $\operatorname{rad}(n)$ is sorted for $1 \le n \le 100000$, find $E(10000)$.

Answer: 21417


The palindromic number $595$ is interesting because it can be written as the sum of consecutive squares: $6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2$.

There are exactly eleven palindromes below one-thousand that can be written as consecutive square sums, and the sum of these palindromes is $4164$. Note that $1 = 0^2 + 1^2$ has not been included as this problem is concerned with the squares of positive integers.

Find the sum of all the numbers less than $10^8$ that are both palindromic and can be written as the sum of consecutive squares.

Answer: 2906969179


The minimum number of cubes to cover every visible face on a cuboid measuring $3 \times 2 \times 1$ is twenty-two.

If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face.

However, the first layer on a cuboid measuring $5 \times 1 \times 1$ also requires twenty-two cubes; similarly the first layer on cuboids measuring $5 \times 3 \times 1$, $7 \times 2 \times 1$, and $11 \times 1 \times 1$ all contain forty-six cubes.

We shall define $C(n)$ to represent the number of cuboids that contain $n$ cubes in one of its layers. So $C(22) = 2$, $C(46) = 4$, $C(78) = 5$, and $C(118) = 8$.

It turns out that $154$ is the least value of $n$ for which $C(n) = 10$.

Find the least value of $n$ for which $C(n) = 1000$.

Answer: 18522


The radical of $n$, $\operatorname{rad}(n)$, is the product of distinct prime factors of $n$. For example, $504 = 2^3 \times 3^2 \times 7$, so $\operatorname{rad}(504) = 2 \times 3 \times 7 = 42$.

We shall define the triplet of positive integers $(a, b, c)$ to be an abc-hit if:

- $\gcd(a, b) = \gcd(a, c) = \gcd(b, c) = 1$

- $a \lt b$

- $a + b = c$

- $\operatorname{rad}(abc) \lt c$

For example, $(5, 27, 32)$ is an abc-hit, because:

- $\gcd(5, 27) = \gcd(5, 32) = \gcd(27, 32) = 1$

- $5 \lt 27$

- $5 + 27 = 32$

- $\operatorname{rad}(4320) = 30 \lt 32$

It turns out that abc-hits are quite rare and there are only thirty-one abc-hits for $c \lt 1000$, with $\sum c = 12523$.

Find $\sum c$ for $c \lt 120000$.

Answer: 18407904


A hexagonal tile with number $1$ is surrounded by a ring of six hexagonal tiles, starting at "12 o'clock" and numbering the tiles $2$ to $7$ in an anti-clockwise direction.

New rings are added in the same fashion, with the next rings being numbered $8$ to $19$, $20$ to $37$, $38$ to $61$, and so on. The diagram below shows the first three rings.

By finding the difference between tile $n$ and each of its six neighbours we shall define $\operatorname{PD}(n)$ to be the number of those differences which are prime.

For example, working clockwise around tile $8$ the differences are $12, 29, 11, 6, 1$, and $13$. So $\operatorname{PD}(8) = 3$.

In the same way, the differences around tile $17$ are $1, 17, 16, 1, 11$, and $10$, hence $\operatorname{PD}(17) = 2$.

It can be shown that the maximum value of $\operatorname{PD}(n)$ is $3$.

If all of the tiles for which $\operatorname{PD}(n) = 3$ are listed in ascending order to form a sequence, the $10$th tile would be $271$.

Find the $2000$th tile in this sequence.

Answer: 14516824220


A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$.

Given that $n$ is a positive integer and $\gcd(n, 10) = 1$, it can be shown that there always exists a value, $k$, for which $R(k)$ is divisible by $n$, and let $A(n)$ be the least such value of $k$; for example, $A(7) = 6$ and $A(41) = 5$.

The least value of $n$ for which $A(n)$ first exceeds ten is $17$.

Find the least value of $n$ for which $A(n)$ first exceeds one-million.

Answer: 1000023


A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$.

Given that $n$ is a positive integer and $\gcd(n, 10) = 1$, it can be shown that there always exists a value, $k$, for which $R(k)$ is divisible by $n$, and let $A(n)$ be the least such value of $k$; for example, $A(7) = 6$ and $A(41) = 5$.

You are given that for all primes, $p \gt 5$, that $p - 1$ is divisible by $A(p)$. For example, when $p = 41$, $A(41) = 5$, and $40$ is divisible by $5$.

However, there are rare composite values for which this is also true; the first five examples being $91$, $259$, $451$, $481$, and $703$.

Find the sum of the first twenty-five composite values of $n$ for which $\gcd(n, 10) = 1$ and $n - 1$ is divisible by $A(n)$.

Answer: 149253


There are some prime values, $p$, for which there exists a positive integer, $n$, such that the expression $n^3 + n^2p$ is a perfect cube.

For example, when $p = 19$, $8^3 + 8^2 \times 19 = 12^3$.

What is perhaps most surprising is that for each prime with this property the value of $n$ is unique, and there are only four such primes below one-hundred.

How many primes below one million have this remarkable property?

Answer: 173


A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$.

For example, $R(10) = 1111111111 = 11 \times 41 \times 271 \times 9091$, and the sum of these prime factors is $9414$.

Find the sum of the first forty prime factors of $R(10^9)$.

Answer: 843296


A number consisting entirely of ones is called a repunit. We shall define $R(k)$ to be a repunit of length $k$; for example, $R(6) = 111111$.

Let us consider repunits of the form $R(10^n)$.

Although $R(10)$, $R(100)$, or $R(1000)$ are not divisible by $17$, $R(10000)$ is divisible by $17$. Yet there is no value of $n$ for which $R(10^n)$ will divide by $19$. In fact, it is remarkable that $11$, $17$, $41$, and $73$ are the only four primes below one-hundred that can  be a factor of $R(10^n)$.

Find the sum of all the primes below one-hundred thousand that will never be a factor of $R(10^n)$.

Answer: 453647705


Consider the consecutive primes $p_1 = 19$ and $p_2 = 23$. It can be verified that $1219$ is the smallest number such that the last digits are formed by $p_1$ whilst also being divisible by $p_2$.

In fact, with the exception of $p_1 = 3$ and $p_2 = 5$, for every pair of consecutive primes, $p_2 \gt p_1$, there exist values of $n$ for which the last digits are formed by $p_1$ and $n$ is divisible by $p_2$. Let $S$ be the smallest of these values of $n$.

Find $\sum S$ for every pair of consecutive primes with $5 \le p_1 \le 1000000$.

Answer: 18613426663617118


Given the positive integers, $x$, $y$, and $z$, are consecutive terms of an arithmetic progression, the least value of the positive integer, $n$, for which the equation, $x^2 - y^2 - z^2 = n$, has exactly two solutions is $n = 27$:
$$34^2 - 27^2 - 20^2 = 12^2 - 9^2 - 6^2 = 27.$$

It turns out that $n = 1155$ is the least value which has exactly ten solutions.

How many values of $n$ less than one million have exactly ten distinct solutions?

Answer: 4989


The positive integers, $x$, $y$, and $z$, are consecutive terms of an arithmetic progression. Given that $n$ is a positive integer, the equation, $x^2 - y^2 - z^2 = n$, has exactly one solution when $n = 20$:
$$13^2 - 10^2 - 7^2 = 20.$$

In fact there are twenty-five values of $n$ below one hundred for which the equation has a unique solution.

How many values of $n$ less than fifty million have exactly one solution?

Answer: 2544559


Consider the infinite polynomial series $A_F(x) = x F_1 + x^2 F_2 + x^3 F_3 + \dots$, where $F_k$ is the $k$th term in the Fibonacci sequence: $1, 1, 2, 3, 5, 8, \dots$; that is, $F_k = F_{k-1} + F_{k-2}$, $F_1 = 1$ and $F_2 = 1$.

For this problem we shall be interested in values of $x$ for which $A_F(x)$ is a positive integer.

| Surprisingly | $\begin{align*}
A_F(\tfrac{1}{2})
&= (\tfrac{1}{2})\times 1 + (\tfrac{1}{2})^2\times 1 + (\tfrac{1}{2})^3\times 2 + (\tfrac{1}{2})^4\times 3 + (\tfrac{1}{2})^5\times 5 + \cdots \\
&= \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{2}{8} + \tfrac{3}{16} + \tfrac{5}{32} + \cdots \\
&= 2
\end{align*}$ |

The corresponding values of $x$ for the first five natural numbers are shown below.

| $x$ | $A_F(x)$ |
| --- | --- |
| $\sqrt{2}-1$ | $1$ |
| $\tfrac{1}{2}$ | $2$ |
| $\frac{\sqrt{13}-2}{3}$ | $3$ |
| $\frac{\sqrt{89}-5}{8}$ | $4$ |
| $\frac{\sqrt{34}-3}{5}$ | $5$ |

We shall call $A_F(x)$ a golden nugget if $x$ is rational, because they become increasingly rarer; for example, the $10$th golden nugget is $74049690$.

Find the $15$th golden nugget.

Answer: 1120149658760


Consider the isosceles triangle with base length, $b = 16$, and legs, $L = 17$.

By using the Pythagorean theorem it can be seen that the height of the triangle, $h = \sqrt{17^2 - 8^2} = 15$, which is one less than the base length.

With $b = 272$ and $L = 305$, we get $h = 273$, which is one more than the base length, and this is the second smallest isosceles triangle with the property that $h = b \pm 1$.

Find $\sum L$ for the twelve smallest isosceles triangles for which $h = b \pm 1$ and $b$, $L$ are positive integers.

Answer: 1118049290473932


Let $(a, b, c)$ represent the three sides of a right angle triangle with integral length sides. It is possible to place four such triangles together to form a square with length $c$.

For example, $(3, 4, 5)$ triangles can be placed together to form a $5$ by $5$ square with a $1$ by $1$ hole in the middle and it can be seen that the $5$ by $5$ square can be tiled with twenty-five $1$ by $1$ squares.

However, if $(5, 12, 13)$ triangles were used then the hole would measure $7$ by $7$ and these could not be used to tile the $13$ by $13$ square.

Given that the perimeter of the right triangle is less than one-hundred million, how many Pythagorean triangles would allow such a tiling to take place?

Answer: 10057761


Consider the infinite polynomial series $A_G(x) = x G_1 + x^2 G_2 + x^3 G_3 + \cdots$, where $G_k$ is the $k$th term of the second order recurrence relation $G_k = G_{k-1} + G_{k-2}$, $G_1 = 1$ and $G_2 = 4$; that is, $1, 4, 5, 9, 14, 23, \dots$.

For this problem we shall be concerned with values of $x$ for which $A_G(x)$ is a positive integer.

The corresponding values of $x$ for the first five natural numbers are shown below.

| $x$ | $A_G(x)$ |
| --- | --- |
| $\frac{\sqrt{5}-1}{4}$ | $1$ |
| $\tfrac{2}{5}$ | $2$ |
| $\frac{\sqrt{22}-2}{6}$ | $3$ |
| $\frac{\sqrt{137}-5}{14}$ | $4$ |
| $\tfrac{1}{2}$ | $5$ |

We shall call $A_G(x)$ a golden nugget if $x$ is rational, because they become increasingly rarer; for example, the $20$th golden nugget is $211345365$.

Find the sum of the first thirty golden nuggets.

Answer: 5673835352990


A positive integer, $n$, is divided by $d$ and the quotient and remainder are $q$ and $r$ respectively. In addition $d$, $q$, and $r$ are consecutive positive integer terms in a geometric sequence, but not necessarily in that order.

For example, $58$ divided by $6$ has quotient $9$ and remainder $4$. It can also be seen that $4, 6, 9$ are consecutive terms in a geometric sequence (common ratio $3/2$).

We will call such numbers, $n$, progressive.

Some progressive numbers, such as $9$ and $10404 = 102^2$, happen to also be perfect squares.
The sum of all progressive perfect squares below one hundred thousand is $124657$.

Find the sum of all progressive perfect squares below one trillion ($10^{12}$).

Answer: 878454337159


Find the smallest $x + y + z$ with integers $x \gt y \gt z \gt 0$ such that $x + y$, $x - y$, $x + z$, $x - z$, $y + z$, $y - z$ are all perfect squares.

Answer: 1006193


Let $ABC$ be a triangle with all interior angles being less than $120$ degrees. Let $X$ be any point inside the triangle and let $XA = p$, $XC = q$, and $XB = r$.

Fermat challenged Torricelli to find the position of $X$ such that $p + q + r$ was minimised.

Torricelli was able to prove that if equilateral triangles $AOB$, $BNC$ and $AMC$ are constructed on each side of triangle $ABC$, the circumscribed circles of $AOB$, $BNC$, and $AMC$ will intersect at a single point, $T$, inside the triangle. Moreover he proved that $T$, called the Torricelli/Fermat point, minimises $p + q + r$. Even more remarkable, it can be shown that when the sum is minimised, $AN = BM = CO = p + q + r$ and that $AN$, $BM$ and $CO$ also intersect at $T$.

If the sum is minimised and $a, b, c, p, q$ and $r$ are all positive integers we shall call triangle $ABC$ a Torricelli triangle. For example, $a = 399$, $b = 455$, $c = 511$ is an example of a Torricelli triangle, with $p + q + r = 784$.

Find the sum of all distinct values of $p + q + r \le 120000$ for Torricelli triangles.

Answer: 30758397


In laser physics, a "white cell" is a mirror system that acts as a delay line for the laser beam. The beam enters the cell, bounces around on the mirrors, and eventually works its way back out.

The specific white cell we will be considering is an ellipse with the equation $4x^2 + y^2 = 100$.

The section corresponding to $-0.01 \le x \le +0.01$ at the top is missing, allowing the light to enter and exit through the hole.

The light beam in this problem starts at the point $(0.0,10.1)$ just outside the white cell, and the beam first impacts the mirror at $(1.4,-9.6)$.

Each time the laser beam hits the surface of the ellipse, it follows the usual law of reflection "angle of incidence equals angle of reflection." That is, both the incident and reflected beams make the same angle with the normal line at the point of incidence.

In the figure on the left, the red line shows the first two points of contact between the laser beam and the wall of the white cell; the blue line shows the line tangent to the ellipse at the point of incidence of the first bounce.

The slope $m$ of the tangent line at any point $(x,y)$ of the given ellipse is: $m = -4x/y$.

The normal line is perpendicular to this tangent line at the point of incidence.

The animation on the right shows the first $10$ reflections of the beam.

How many times does the beam hit the internal surface of the white cell before exiting?

Answer: 354


Some positive integers $n$ have the property that the sum $[n + \operatorname{reverse}(n)]$ consists entirely of odd (decimal) digits. For instance, $36 + 63 = 99$ and $409 + 904 = 1313$. We will call such numbers reversible; so $36$, $63$, $409$, and $904$ are reversible. Leading zeroes are not allowed in either $n$ or $\operatorname{reverse}(n)$.

There are $120$ reversible numbers below one-thousand.

How many reversible numbers are there below one-billion ($10^9$)?

Answer: 608720


The smallest positive integer $n$ for which the numbers $n^2 + 1$, $n^2 + 3$, $n^2 + 7$, $n^2 + 9$, $n^2 + 13$, and $n^2 + 27$ are consecutive primes is $10$. The sum of all such integers $n$ below one-million is $1242490$.

What is the sum of all such integers $n$ below $150$ million?

Answer: 676333270


In a $3 \times 2$ cross-hatched grid, a total of $37$ different rectangles could be situated within that grid as indicated in the sketch.

There are $5$ grids smaller than $3 \times 2$, vertical and horizontal dimensions being important, i.e. $1 \times 1$, $2 \times 1$, $3 \times 1$, $1 \times 2$ and $2 \times 2$. If each of them is cross-hatched, the following number of different rectangles could be situated within those smaller grids:

| $1 \times 1$ | $1$ |
| $2 \times 1$ | $4$ |
| $3 \times 1$ | $8$ |
| $1 \times 2$ | $4$ |
| $2 \times 2$ | $18$ |

Adding those to the $37$ of the $3 \times 2$ grid, a total of $72$ different rectangles could be situated within $3 \times 2$ and smaller grids.

How many different rectangles could be situated within $47 \times 43$ and smaller grids?

Answer: 846910284


We can easily verify that none of the entries in the first seven rows of Pascal's triangle are divisible by $7$:

|  |  |  |  |  |  | $1$ |  |  |  |  |  |  |
|  |  |  |  |  | $1$ |  | $1$ |  |  |  |  |  |
|  |  |  |  | $1$ |  | $2$ |  | $1$ |  |  |  |  |
|  |  |  | $1$ |  | $3$ |  | $3$ |  | $1$ |  |  |  |
|  |  | $1$ |  | $4$ |  | $6$ |  | $4$ |  | $1$ |  |  |
|  | $1$ |  | $5$ |  | $10$ |  | $10$ |  | $5$ |  | $1$ |  |
| $1$ |  | $6$ |  | $15$ |  | $20$ |  | $15$ |  | $6$ |  | $1$ |

However, if we check the first one hundred rows, we will find that only $2361$ of the $5050$ entries are not divisible by $7$.

Find the number of entries which are not divisible by $7$ in the first one billion ($10^9$) rows of Pascal's triangle.

Answer: 2129970655314432


Looking at the table below, it is easy to verify that the maximum possible sum of adjacent numbers in any direction (horizontal, vertical, diagonal or anti-diagonal) is $16$ ($= 8 + 7 + 1$).

| $-2$ | $5$ | $3$ | $2$ |
| $9$ | $-6$ | $5$ | $1$ |
| $3$ | $2$ | $7$ | $3$ |
| $-1$ | $8$ | $-4$ | $8$ |

Now, let us repeat the search, but on a much larger scale:

First, generate four million pseudo-random numbers using a specific form of what is known as a "Lagged Fibonacci Generator":

For $1 \le k \le 55$, $s_k = [100003 - 200003 k + 300007 k^3] \pmod{1000000} - 500000$.

For $56 \le k \le 4000000$, $s_k = [s_{k-24} + s_{k - 55} + 1000000] \pmod{1000000} - 500000$.

Thus, $s_{10} = -393027$ and $s_{100} = 86613$.

The terms of $s$ are then arranged in a $2000 \times 2000$ table, using the first $2000$ numbers to fill the first row (sequentially), the next $2000$ numbers to fill the second row, and so on.

Finally, find the greatest sum of (any number of) adjacent entries in any direction (horizontal, vertical, diagonal or anti-diagonal).

Answer: 52852124