Four-digit numbers whose digits sum to 9: every AI model's reply, tested

How many four-digit positive integers have digits that add up to 9?
Show your working briefly, then end with a line that says "Final answer: " followed by the answer alone.

Exactly what this prompt's replies are checked against, with every other prompt of our test runs.

1. GPT-6 Luna

   Passed: Final answer 165: right.

   Let the first digit be at least 1. Subtract 1 from it, so the four digits become nonnegative integers summing to 8. By stars and bars, there are \(\binom{8+4-1}{4-1}=\binom{11}{3}=165\) possibilities.
   
   Final answer: 165

   378 tokens in, 127 out (50 of them reasoning) · 2.0 s · $0.0001 · 1 message on Pro · answered by openai/gpt-6-luna via OpenAI · September 27, 2026

2. GLM 5.3 Flash

   Passed: Final answer 165: right.

   We need digits d₁+d₂+d₃+d₄ = 9 with d₁ ≥ 1 (thousands digit), others 0–9 (no digit can exceed 9 since the sum is 9, so that constraint is automatic).
   
   Substitute e₁ = d₁ − 1 ≥ 0: e₁+d₂+d₃+d₄ = 8.
   
   Number of nonnegative solutions: C(8+4−1, 4−1) = C(11,3) = 165.
   
   Final answer: 165

   393 tokens in, 171 out (54 of them reasoning) · 2.7 s · $0.0001 · 1 message on Pro · answered by z-ai/glm-5.3-flash via Together · September 27, 2026

3. Claude Haiku 5.5

   Passed: Final answer 165: right.

   We need four-digit numbers d1 d2 d3 d4 with d1 from 1 to 9, d2, d3, d4 from 0 to 9, and digit sum 9.
   
   Substitute e1 = d1 − 1, so e1 ranges 0 to 8, and e1 + d2 + d3 + d4 = 8 with all variables from 0 to 9.
   
   Since the sum is 8, no variable can exceed 8, so the upper bound of 9 is never violated. The number of nonnegative solutions is C(8+3, 3) = C(11, 3) = 165.
   
   Check: total solutions with d1 ≥ 0 allowed (i.e., first digit could be 0) for sum 9 across four digits is C(12, 3) = 220. Those with d1 = 0 correspond to three digits summing to 9: C(11, 2) = 55. So 220 − 55 = 165, which matches.
   
   Final answer: 165

   513 tokens in, 305 out · 1.9 s · $0.0002 · 1 message on Pro · answered by anthropic/claude-haiku-5.5 via Anthropic · October 7, 2026

4. GLM 5.3

   Passed: Final answer 165: right.

   # Working
   
   We need four-digit numbers (1000–9999) with digit sum 9.
   
   Let the digits be $a, b, c, d$ where $a \in \{1,\dots,9\}$ and $b,c,d \in \{0,\dots,9\}$, with $a+b+c+d=9$.
   
   Set $a' = a-1 \geq 0$, so $a'+b+c+d = 8$ with each variable $\leq 9$ (automatically satisfied since sum is 8).
   
   Stars and bars: $\binom{8+3}{3} = \binom{11}{3} = 165$.
   
   Final answer: 165

   392 tokens in, 154 out · 1.9 s · $0.0003 · 1 message on Pro · answered by z-ai/glm-5.3 via Baidu · September 27, 2026

5. DeepSeek V4.1 Flash

   Passed: Final answer 165: right.

   Let the digits be a, b, c, d with a ≥ 1 (it's a four-digit number) and b, c, d ≥ 0, and a+b+c+d = 9.
   
   Substitute a = a' + 1 where a' ≥ 0:
   a' + b + c + d = 8
   
   Number of non-negative solutions = C(8+3, 3) = C(11, 3) = 165.
   
   Final answer: 165

   417 tokens in, 197 out (92 of them reasoning) · 0.7 s · $0.0003 · 1 message on Pro · answered by deepseek/deepseek-v4.1-flash via Together · October 8, 2026

6. Mistral Large 4

   Passed: Final answer 165: right.

   Let the four-digit number be \(abcd\), where \(a \ge 1\) and \(b,c,d \ge 0\). We need
   
   \[
   a+b+c+d=9.
   \]
   
   Set \(a'=a-1\), so \(a',b,c,d \ge 0\). Then
   
   \[
   a'+b+c+d=8.
   \]
   
   By stars and bars, the number of nonnegative solutions is
   
   \[
   \binom{8+4-1}{4-1}=\binom{11}{3}=165.
   \]
   
   Final answer: 165

   383 tokens in, 219 out (72 of them reasoning) · 2.6 s · $0.0007 · 1 message on Pro · answered by mistralai/mistral-large-4-0 via Mistral · October 9, 2026

7. GPT-6.1 Sol

   Passed: Final answer 165: right.

   Let the digits be \(a,b,c,d\), with \(a\ge1\). Set \(x=a-1\). Then
   \[
   x+b+c+d=8,
   \]
   where all four variables are nonnegative. Each resulting digit is automatically at most 9.
   
   By stars and bars, the number of solutions is
   \[
   \binom{8+4-1}{4-1}=\binom{11}{3}=165.
   \]
   
   Final answer: 165

   380 tokens in, 147 out (39 of them reasoning) · 2.7 s · $0.0011 · 1 message on Pro · answered by openai/gpt-6.1-sol via OpenAI · September 29, 2026

8. DeepSeek V4 Pro

   Passed: Final answer 165: right.

   Let the four digits be \(a,b,c,d\), where \(a \ge 1\) and the others are nonnegative. We need:
   
   \[
   a+b+c+d=9
   \]
   
   Set \(a'=a-1\), so \(a' \ge 0\). Then:
   
   \[
   a'+b+c+d=8
   \]
   
   The number of nonnegative integer solutions is:
   
   \[
   \binom{8+4-1}{4-1}=\binom{11}{3}=165
   \]
   
   Final answer: 165

   389 tokens in, 269 out (160 of them reasoning) · 4.0 s · $0.0012 · 1 message on Pro · answered by deepseek/deepseek-v4-pro-0813 via Wafer · October 8, 2026

9. GPT-6 Sol

   Passed: Final answer 165: right.

   Subtract 1 from the thousands digit. The four digits then become nonnegative integers that sum to 8. By stars and bars, the number of solutions is \(\binom{8+4-1}{4-1}=\binom{11}{3}=165\). Every solution gives a valid four-digit integer.
   
   Final answer: 165

   378 tokens in, 117 out (39 of them reasoning) · 2.5 s · $0.0019 · 1 message on Pro · answered by openai/gpt-6-sol via OpenAI · September 27, 2026

10. Gemini 4 Argon

    Passed: Final answer 165: right.

    Let the four-digit integer be represented by its digits $d_1, d_2, d_3, d_4$. We want to find the number of solutions to the equation:
    $$d_1 + d_2 + d_3 + d_4 = 9$$
    where $d_1 \ge 1$ and $d_2, d_3, d_4 \ge 0$. (Since the maximum sum is 9, no digit can exceed 9, so we do not need to worry about the upper bound of $\le 9$ for each digit.)
    
    Let $d_1' = d_1 - 1$, so that $d_1' \ge 0$. Substituting this into the equation gives:
    $$d_1' + d_2 + d_3 + d_4 = 8$$
    
    Using the stars and bars method for $n = 8$ and $k = 4$, the number of non-negative integer solutions is:
    $$\binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$$
    
    Final answer: 165

    623 tokens in, 284 out · 4.5 s · $0.0020 · 1 message on Pro · answered by google/gemini-4-argon via Google AI Studio · October 15, 2026

11. Gemini 3.8 Flash

    Passed: Final answer 165: right.

    Let the four digits be represented by $a, b, c,$ and $d$, where $a \in \{1, 2, \dots, 9\}$ and $b, c, d \in \{0, 1, \dots, 9\}$.
    
    We need to find the number of integer solutions to:
    $$a + b + c + d = 9$$
    
    Let $a' = a - 1$ so that $a' \ge 0$. Substituting gives:
    $$a' + b + c + d = 8$$
    
    Since the sum is $8$, no individual variable can exceed $8$, so the constraint that digits must be $\le 9$ is automatically satisfied.
    
    Using the stars and bars formula, the number of non-negative integer solutions is:
    $$\binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165$$
    
    Final answer: 165

    404 tokens in, 495 out (260 of them reasoning) · 4.7 s · $0.0022 · 1 message on Pro · answered by google/gemini-3.8-flash via Google · September 27, 2026

12. Claude Haiku 4.5

    Passed: Final answer 165: right.

    I need to find four-digit positive integers where the digits sum to 9.
    
    A four-digit positive integer has the form $\overline{abcd}$ where:
    - $a \in \{1,2,...,9\}$ (first digit can't be 0)
    - $b,c,d \in \{0,1,...,9\}$
    - $a + b + c + d = 9$
    
    **Method: Change of variables**
    
    Let $a' = a - 1$, so $a' \in \{0,1,...,8\}$.
    
    Then: $a' + 1 + b + c + d = 9$
    
    This gives: $a' + b + c + d = 8$
    
    Now I need to count non-negative integer solutions where:
    - $a' \in \{0,1,...,8\}$
    - $b,c,d \in \{0,1,...,9\}$
    - $a' + b + c + d = 8$
    
    Since the sum is only 8, each variable is automatically ≤ 8 (and thus $a' \leq 8$ and $b,c,d \leq 9$ are automatically satisfied).
    
    This is a standard stars and bars problem: distribute 8 identical objects into 4 distinct bins with no restrictions.
    
    The number of solutions is:
    $$\binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165$$
    
    Final answer: 165

    404 tokens in, 404 out · 2.9 s · $0.0024 · 1 message on Pro · answered by anthropic/claude-haiku-4.5 via Amazon Bedrock · September 27, 2026

13. Claude Sonnet 5.5

    Passed: Final answer 165: right.

    Let the digits be a, b, c, d with a ≥ 1 and a + b + c + d = 9.
    
    Substitute a' = a − 1 ≥ 0, so a' + b + c + d = 8 with all variables ≥ 0. (Digits can't exceed 9, and since the sum is 8 or 9, no digit exceeds 9 anyway.)
    
    Number of non-negative solutions: C(8+3, 3) = C(11, 3) = 165.
    
    Final answer: 165

    514 tokens in, 161 out · 1.8 s · $0.0026 · 1 message on Pro · answered by anthropic/claude-sonnet-5.5 via Anthropic · September 28, 2026

14. Kimi K3

    Passed: Final answer 165: right.

    Let the digits be \(a,b,c,d\) with \(a\ge1\) and \(a+b+c+d=9\).
    
    Set \(a'=a-1\), so \(a'\ge0\) and:
    
    \[
    a'+b+c+d=8
    \]
    
    Since the total is 8, no digit can exceed 9, so upper limits don’t matter. By stars and bars:
    
    \[
    \binom{8+4-1}{4-1}=\binom{11}{3}=165
    \]
    
    Final answer: 165

    465 tokens in, 332 out (210 of them reasoning) · 6.0 s · $0.0032 · 1 message on Pro · answered by moonshotai/kimi-k3 via Wafer · September 27, 2026

15. Claude Sonnet 5

    Passed: Final answer 165: right.

    Let digits be d1 d2 d3 d4, with d1≥1 (leading digit), all digits 0-9, sum = 9.
    
    Substitute d1' = d1 - 1 ≥ 0, so d1' + d2 + d3 + d4 = 8, all ≥0.
    
    Since max possible value (8) is less than 10, no digit can exceed 9, so no upper-bound restrictions apply.
    
    Number of nonnegative integer solutions: C(8+3, 3) = C(11,3) = 165.
    
    Final answer: 165

    510 tokens in, 290 out (111 of them reasoning) · 4.5 s · $0.0039 · 1 message on Pro · answered by anthropic/claude-sonnet-5 via Claude Platform on AWS · September 27, 2026

16. Claude Opus 5.5

    Passed: Final answer 165: right.

    Let the number have digits $a, b, c, d$ with $a \ge 1$ and $a + b + c + d = 9$.
    
    Substitute $a' = a - 1 \ge 0$. Then $a' + b + c + d = 8$, where all variables are non-negative.
    
    By stars and bars, the number of solutions is $\binom{8+3}{3} = \binom{11}{3} = 165$.
    
    No digit can exceed 9, because the largest possible value is $a = 9$ (when $a' = 8$). So every solution gives a valid four-digit number.
    
    Final answer: 165

    512 tokens in, 282 out (91 of them reasoning) · 4.7 s · $0.0077 · 1 message on Pro · answered by anthropic/claude-opus-5.5 via Claude Platform on AWS · September 27, 2026

17. Grok 4.7

    Passed: Final answer 165: right.

    The four-digit positive integers whose digits sum to 9 are the solutions of \(A + B + C + D = 9\) in which \(A \in \{1,\dots,9\}\) and \(B,C,D \in \{0,\dots,9\}\).
    
    Because the target sum is only 9, the upper bounds of 9 are automatically satisfied. The change of variables \(A' = A - 1\) (so \(A' \ge 0\)) therefore converts the problem into the unrestricted equation
    \[
    A' + B + C + D = 8
    \]
    in non-negative integers. The number of solutions is
    \[
    \binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} = 165.
    \]
    
    Final answer: 165

    1,614 tokens in, 1,305 out (1,127 of them reasoning) · 15.1 s · $0.0093 · 1 message on Pro · answered by x-ai/grok-4.7 via xAI · October 2, 2026

18. GPT-6 Astra

    Passed: Final answer 165: right.

    Let the digits be \(a,b,c,d\), where \(a\ge1\) and the others are nonnegative. We need
    \[
    a+b+c+d=9.
    \]
    Set \(x=a-1\). Then \(x+b+c+d=8\). By stars and bars, the number of nonnegative solutions is
    \[
    \binom{8+4-1}{4-1}=\binom{11}{3}=165.
    \]
    All solutions give valid digits, since \(a\le9\) and \(b,c,d\le8\).
    
    Final answer: 165

    378 tokens in, 169 out (35 of them reasoning) · 3.8 s · $0.0122 · 1 message on Pro · answered by openai/gpt-6-astra via OpenAI · September 27, 2026

19. Claude Fable 5.1

    Passed: Final answer 165: right.

    Let the digits be a, b, c, d with a ≥ 1 (leading digit) and b, c, d ≥ 0, and a + b + c + d = 9.
    
    Substitute a' = a − 1 ≥ 0, so a' + b + c + d = 8 with all variables ≥ 0.
    
    Number of non-negative integer solutions: C(8 + 3, 3) = C(11, 3) = 165.
    
    No digit can exceed 9 since the total is only 9, so no solutions need to be excluded.
    
    Final answer: 165

    512 tokens in, 175 out · 4.6 s · $0.0139 · 1 message on Pro · answered by anthropic/claude-fable-5.1 via Anthropic · September 27, 2026

20. Gemini 3.1 Pro

    Passed: Final answer 165: right.

    Let the four-digit positive integer be represented by its digits as $abcd$. We are given that $a + b + c + d = 9$. 
    Since it is a four-digit number, the leading digit $a$ must be at least 1 ($1 \le a \le 9$), while the other digits can be any integer from 0 to 9.
    
    We can use the substitution $x = a - 1$, where $x \ge 0$. 
    Substituting $a = x + 1$ into our equation gives:
    $(x + 1) + b + c + d = 9$
    $x + b + c + d = 8$
    
    We now need to find the number of non-negative integer solutions to this equation. Using the "stars and bars" combinatorial method, the number of solutions for distributing $n$ items among $k$ distinct bins is given by the binomial coefficient $\binom{n + k - 1}{k - 1}$.
    
    Here, $n = 8$ and $k = 4$ (the variables $x, b, c, d$). Since the sum is 8, none of the variables will exceed 9, so no upper limit restrictions are violated.
    Number of solutions = $\binom{8 + 4 - 1}{4 - 1} = \binom{11}{3}$
    
    Calculating $\binom{11}{3}$:
    $\frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165$
    
    Final answer: 165

    404 tokens in, 1,181 out (826 of them reasoning) · 9.6 s · $0.0150 · 1 message on Pro · answered by google/gemini-3.1-pro-preview via Google · September 27, 2026

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