---
library_name: transformers
license: apache-2.0
base_model:
- Qwen/Qwen3.5-9B
base_model_relation: finetune
pipeline_tag: image-text-to-text
language:
- en
tags:
- physics
- science
- reasoning
- math
- qwen3.5
---
# Quark
Quark is a 9B reasoning model for physics, science and maths, built by [FirstPrinciples](https://firstprinciples.com/). The base model is
[Qwen3.5-9B](https://huggingface.co/Qwen/Qwen3.5-9B) with additional post-training: supervised fine-tuning on
our curated physics, maths and coding mix.
It outperforms the base model on a range of benchmarks, and reaches
graduate-level science scores comparable to much larger open models. We also quantized the model weights to 8 bits (Q8_0) for llama.cpp, Ollama and LM Studio; this
[Q8_0 version](#q8_0-version) matches bf16 on every benchmark below.
Quark is the first in a series of small, powerful scientific reasoning models. You can run it locally (see [Usage](#usage)) or on [Theo Platform](https://firstprinciples.com/sign-up), a research platform for complex problems, where you can use it alongside other frontier models, check its results with Python, Mathematica and Sage, and keep each session as a research graph you can revise one step at a time.
## Results
Accuracy (%), measured with [EvalScope](https://github.com/modelscope/evalscope) and vLLM (the Q8_0 version
with llama.cpp): thinking mode, up to 131,072 output tokens, and following the Qwen3.5 model card's sampling
settings. AIME 2025 is averaged over 16 samples per problem and GPQA Diamond over 8. HLE (text-only, 2,158
questions) is graded by Qwen3.5-27B as judge.
| Benchmark | Qwen3.5-9B | Quark (bf16) | Quark ([Q8_0](#q8_0-version)) |
| --- | --- | --- | --- |
| GPQA Diamond | 81.6 | 87.4 | **87.8** |
| AIME 2025 | 86.9 | 91.7 | **94.0** |
| MMLU-Pro Physics | 90.7 | 91.0 | **91.6** |
| Humanity's Last Exam (HLE), text | 15.9 | **22.7** | 22.3 |
| HLE, physics subset | 8.4 | 15.8 | **16.3** |
All were measured by us with the same setup.
**Compared with other open models** (scores as reported by each developer):
| Model | Size | GPQA Diamond | AIME 2025 | HLE |
| --- | --- | --- | --- | --- |
| Quark (fp16) | 9B | 87.4 | 91.7 | 22.7 |
| Quark (Q8_0) | 9B | 87.8 | 94.0 | 22.3 |
| Qwen3-8B | 8B | 59.6 | 69.3 | 4.4 |
| NVIDIA Nemotron Nano 9B v2 | 9B | 64.0 | 72.1 | 6.5 |
| gpt-oss-20b | 21B | 71.5 | 91.7 | 10.9 |
| Qwen3.5-27B | 27B | 85.5 | — | 24.3 |
| Qwen3.8-27B | 27B | **89.2** | — | **30.8** |
| gpt-oss-120b | 117B | 80.1 | 92.5 | 14.9 |
| DeepSeek-R1-0528 | 671B | 81.0 | 87.5 | 17.7 |
| Kimi K2 Thinking | 1T | 84.5 | **94.5** | 23.9 |
— = not reported. Sources: each model's official model card; Quark (bf16) and its Q8_0 version measured by us
(see above)
## Q8_0 version
A Q8_0 version (model weights quantized to 8 bits) for llama.cpp, Ollama and LM Studio (in their GGUF format) is available at
[FirstPrinciplesLabs/quark-GGUF](https://huggingface.co/FirstPrinciplesLabs/quark-GGUF). Its accuracy is
in the [Results](#results) table: it matches Quark on all benchmarks.
## Throughput
Tokens per second on an NVIDIA H200 and a MacBook Pro, for the original bf16 weights and the
[Q8_0 version](#q8_0-version).
### llama.cpp
Throughput (token/s) for a single user with llama.cpp, by how much context is already filled.
| Hardware | Context | Q8_0 | bf16 |
| --- | --- | --- | --- |
| NVIDIA H200 | empty | **216** | 185 |
| NVIDIA H200 | 32K | **191** | 167 |
| NVIDIA H200 | 100K | **156** | 139 |
| MacBook Pro, M4, 16 GB | empty | **11.2** | does not fit (18.4 GB) |
| MacBook Pro, M4, 16 GB | 32K | **8.7** | does not fit |
## Training data
Quark was fine-tuned on our curated mix of about 200,000 physics, science, maths and coding problems,
each with a worked solution written by [Qwen3.8-Flash-Next](https://huggingface.co/Qwen/Qwen3.8-Flash-Next).
Solutions were checked for correctness, and problems overlapping the benchmarks above were removed.
## Usage
Quark runs anywhere Qwen3.5-9B does (Transformers, vLLM, SGLang). We recommend an output budget of 131,072 tokens for best performance.
```bash
vllm serve FirstPrinciplesLabs/quark \
--max-model-len 135168 --reasoning-parser qwen3
```
Or run it locally with [Ollama](https://ollama.com/FirstPrinciples/quark) (BF16 weights, with the recommended sampling settings and a 131,072-token context built in):
```bash
ollama run firstprinciples/quark
```
For the [Q8_0 version](#q8_0-version) with llama.cpp:
```bash
llama-server -hf FirstPrinciplesLabs/quark-GGUF:Q8_0 --jinja -c 131072 \
--temp 1.0 --top-k 20 --top-p 0.95 --presence-penalty 1.5
```
Recommended sampling (from the Qwen3.5 model card):
| | Thinking (default) | Non-thinking |
| --- | --- | --- |
| `chat_template_kwargs` | `{"enable_thinking": true}` | `{"enable_thinking": false}` |
| temperature / top_p / top_k | 1.0 / 0.95 / 20 | 1.0 / 1.0 / 40 |
| presence penalty | 1.5 | 2.0 |
| max output tokens | up to 131,072 | 32,768 |
## Example
A hard problem from AIME 2025 (II, problem 15), in thinking mode with the recommended sampling settings:
> Let \\(f(x)=\dfrac{(x-18)(x-72)(x-98)(x-k)}{x}\\).
> There exist exactly three positive real values of \\(k\\) such that \\(f\\) has a minimum at exactly two real values of \\(x\\). Find the sum of these three values of \\(k\\).
| | Qwen3.5-9B | **Quark** |
| --- | --- | --- |
| Correct, out of 16 attempts | 1 | **10** |
| Most common answer | 188 ✗ (14 of 16 attempts) | **240 ✓** |
| Approach | Finds no way in and guesses that k must be one of the existing roots: 18 + 72 + 98 = 188 | Turns "a minimum at two points" into an equation for k and solves it |
**How Quark solves it** (summarised from its answer):
1. If the minimum value \\(m\\) is reached at two points \\(u\\) and \\(v\\), the numerator minus \\(mx\\) must be a perfect square:
$$
(x-18)(x-72)(x-98)(x-k)-mx=(x-u)^2(x-v)^2 .
$$
1. Matching coefficients and eliminating \\(u\\) and \\(v\\) leaves one equation in \\(k\\):
$$
(k^2-376k-5120)^2 = 8128512\,k .
$$
1. It factors as \\((k-8)(k-32)(k-200)(k-512)=0\\), and checking the sign before squaring rules out \\(k=512\\).
2. So \\(k = 8, 32, 200\\), and the answer is **240**.
Both models' full written answers (each follows several thousand tokens of reasoning, not shown):
Quark: 240 (correct)
We interpret “ \\(f\\) has a minimum at exactly two real values of \\(x\\)” as meaning that the minimum value is attained at exactly two distinct \\(x\\)-values (the intended interpretation; otherwise \\(f\\) has no global minimum on all real \\(x\neq0\\)).
Let
$$
P(x)=(x-18)(x-72)(x-98)(x-k).
$$
First expand \\(P(x)\\). We have
$$
18+72+98=188,
$$
$$
18\cdot72+18\cdot98+72\cdot98=10116,
$$
$$
18\cdot72\cdot98=127008.
$$
Thus
$$
P(x)=x^4-(k+188)x^3+(10116+188k)x^2-(127008+10116k)x+127008k.
$$
Suppose the minimum value is \\(m\\), attained at two distinct \\(x\\)-values \\(u\neq v\\). Then \\(f(x)-m\\) has double zeros at \\(u\\) and \\(v\\). Since
$$
x(f(x)-m)=P(x)-mx,
$$
we must have
$$
P(x)-mx=(x-u)^2(x-v)^2.
$$
Now expand the right-hand side:
$$
(x-u)^2(x-v)^2
=
x^4-2(u+v)x^3+\bigl((u+v)^2+2uv\bigr)x^2-2uv(u+v)x+u^2v^2.
$$
Matching coefficients gives
$$
2(u+v)=k+188, \tag{1}
$$
$$
(u+v)^2+2uv=10116+188k, \tag{2}
$$
$$
u^2v^2=127008k. \tag{3}
$$
Let
$$
r=u+v,\qquad s=uv.
$$
Since \\(u,v\gt 0\\), we have \\(r\gt 0\\) and \\(s\gt 0\\). From (1),
$$
r=\frac{k+188}{2}.
$$
From (3),
$$
s=\sqrt{127008k}.
$$
Substitute into (2):
$$
\left(\frac{k+188}{2}\right)^2+2\sqrt{127008k}=10116+188k.
$$
Simplify:
$$
\frac{k^2}{4}+94k+8836+2\sqrt{127008k}=10116+188k,
$$
$$
\frac{k^2}{4}-94k-1280=-2\sqrt{127008k}.
$$
Multiplying by \\(4\\),
$$
k^2-376k-5120=-8\sqrt{127008k}. \tag{4}
$$
Now square both sides:
$$
(k^2-376k-5120)^2=64\cdot127008\,k.
$$
Since \\(64\cdot127008=8128512\\),
$$
(k^2-376k-5120)^2-8128512k=0.
$$
This factors as
$$
(k-8)(k-32)(k-200)(k-512)=0.
$$
But equation (4) requires the left-hand side \\(k^2-376k-5120\\) to be negative. Checking the candidates:
$$
\begin{aligned}
k=8&:\quad 64-3008-5120=-8064,\\
k=32&:\quad 1024-12032-5120=-16128,\\
k=200&:\quad 40000-75200-5120=-40320,\\
k=512&:\quad 262144-192512-5120=64512.
\end{aligned}
$$
Thus \\(k=512\\) is extraneous, while
$$
k=8,\;32,\;200
$$
are valid.
Their sum is
$$
8+32+200=240.
$$
$$
\boxed{240}
$$
Qwen3.5-9B: 188 (incorrect)
The function is given by \\(f(x)=\frac{(x-18)(x-72)(x-98)(x-k)}{x}\\).
We are looking for the number of local minima.
Let \\(P(x) = (x-18)(x-72)(x-98)(x-k)\\). Then \\(f(x) = P(x)/x\\).
The critical points of \\(f(x)\\) are given by \\(f'(x) = 0\\), which is equivalent to \\(x P'(x) - P(x) = 0\\). Let \\(Q(x) = x P'(x) - P(x)\\).
\\(Q(x)\\) is a polynomial of degree 4.
We analyzed the signs of \\(Q(x)\\) at the roots of \\(P(x)\\), denoted \\(r_1, r_2, r_3, r_4\\) (sorted).
We found that \\(Q(r_i)\\) alternates in sign: \\(Q(r_1) \lt 0\\), \\(Q(r_2) \gt 0\\), \\(Q(r_3) \lt 0\\), \\(Q(r_4) \gt 0\\).
Also \\(Q(0) \lt 0\\).
Using the Intermediate Value Theorem, \\(Q(x)\\) has roots in the intervals \\((-\infty, 0)\\), \\((r_1, r_2)\\), \\((r_2, r_3)\\), and \\((r_3, r_4)\\).
Let these roots be \\(c_1, c_2, c_3, c_4\\).
\\(c_1\\) is in \\((-\infty, 0)\\), and corresponds to a local maximum (since \\(f \to -\infty\\) at both ends of this interval).
\\(c_2\\) is in \\((r_1, r_2)\\), and corresponds to a local minimum (since \\(f\\) changes from 0 to negative to 0).
\\(c_3\\) is in \\((r_2, r_3)\\), and corresponds to a local maximum (since \\(f\\) changes from 0 to positive to 0).
\\(c_4\\) is in \\((r_3, r_4)\\), and corresponds to a local minimum (since \\(f\\) changes from 0 to negative to 0).
Thus, for generic \\(k\\) (distinct roots), \\(f(x)\\) has exactly two local minima.
The problem states there are exactly three values of \\(k\\). This suggests that for these values, the condition holds, or perhaps the interpretation of "minimum" or the number of extrema changes. However, based on standard interpretations of such problems where the set of solutions is finite and associated with the roots of the polynomial, the values of \\(k\\) are likely the roots of \\(P(x)\\) where the polynomial has a multiple root.
If \\(k\\) coincides with one of the existing roots \\(18, 72, 98\\), the polynomial \\(P(x)\\) has a root of multiplicity at least 2.
Specifically, if \\(k=18\\), \\(r_1=18\\) is a double root. If \\(k=72\\), \\(r_2=72\\) is a double root. If \\(k=98\\), \\(r_3=98\\) is a double root.
These values are natural candidates.
If \\(k\\) is distinct from 18, 72, 98, and positive, the number of minima is consistently found to be 2.
Given the constraint "exactly three positive real values", it implies that the condition holds specifically for \\(k=18, 72, 98\\).
(Note: There might be a subtle nuance regarding the definition of minimum or global minimum, or perhaps my analysis of generic \\(k\\) having 2 minima implies the condition holds for all \\(k\\), but the problem structure implies a finite set. In competitions, such "exactly three" questions often point to the roots).
Assuming the values are \\(18, 72, 98\\).
Sum \\(= 18 + 72 + 98 = 188\\).
\\(\boxed{188}\\)
## Limitations
- **Needs a long output budget.** At a 32K-token limit about 20% of GPQA answers are cut off and GPQA drops
to 74.9, below the base model.
- **Non-thinking mode is slightly weaker than the base model** (AIME 56.7 vs 59.8, GPQA 76.8 vs 79.6).
- **Specialised:** trained only on single-turn science, maths and code questions; not trained for
conversation, tool use, images or safety beyond what Qwen3.5-9B already has.
## License
Apache 2.0