| --- |
| source: ncert |
| topic: numbers |
| class_level: class_8 |
| chapter: ch16 |
| difficulty: intermediate |
| --- |
| |
| Class 8 | Ch16: Playing With Numbers |
| NUMBERS IN GENERAL FORM: |
| 2-digit number ab = 10a+b (a=tens digit, b=units digit). |
| 3-digit number abc = 100a+10b+c. |
| Example: 57 = 10×5+7. 347 = 100×3+10×4+7. |
| REVERSING DIGITS: |
| 2-digit number ab=10a+b. Reversed ba=10b+a. |
| Sum: (10a+b)+(10b+a)=11(a+b). Always divisible by 11. |
| Difference: (10a+b)-(10b+a)=9(a-b). Always divisible by 9. |
| 3-digit number abc=100a+10b+c. Reversed cba=100c+10b+a. |
| Difference: (100a+10b+c)-(100c+10b+a)=99(a-c). Always divisible by 99. |
| CYCLIC NUMBERS (3 digits abc, bca, cab): |
| abc+bca+cab=111(a+b+c). Always divisible by 111 and 3. |
| LETTERS FOR DIGITS (cryptarithmetic): |
| Each letter represents a unique digit. Use logic to find values. |
| AB+BA=121 → (10a+b)+(10b+a)=11(a+b)=121 → a+b=11. |
| DIVISIBILITY TESTS REVISITED: |
| Div by 3: sum of digits divisible by 3. |
| Div by 9: sum of digits divisible by 9. |
| Div by 11: (sum of odd position digits)-(sum of even position digits)=0 or multiple of 11. |
| SOLVED EXAMPLES: |
| Example 1: A 2-digit number is 4 times sum of its digits. If 18 added, digits reverse. |
| Let number=10a+b. 10a+b=4(a+b) → 10a+b=4a+4b → 6a=3b → b=2a. |
| After adding 18: 10a+b+18=10b+a → 9a-9b=-18 → b-a=2. |
| From b=2a and b-a=2: 2a-a=2 → a=2, b=4. Number=24. |
| Verify: 24=4×(2+4)=4×6=24. 24+18=42 (digits reversed). Correct. |
| Example 2: If 1A×A=9A where A is same digit, find A. |
| Try A=5: 15×5=75=9×5? No. Try A=6: 16×6=96=9×6? 9×6=54. No. |
| Think: 1A×A means (10+A)×A=10A+A^2. This should equal 90+A. |
| 10A+A^2=90+A → A^2+9A-90=0 → (A+15)(A-6)=0 → A=6. |
| Check: 16×6=96. Yes! 1A×A=16×6=96=9A where A=6. |
| COMMON MISTAKES: |
| General form of 2-digit number ab is 10a+b NOT a×b or a+b. |
| When digits are reversed: the tens digit becomes units and vice versa. |
