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. DeepSeek V4.1 Flash

   Passed: Final answer 165: right.

   Let the digits be $a,b,c,d$ with $a \ge 1$ (four-digit) and $b,c,d \ge 0$, and $a+b+c+d=9$.
   
   Set $a' = a-1 \ge 0$. Then $a'+b+c+d = 8$.
   
   Number of nonnegative solutions to a sum of 8 across 4 variables:
   
   $$\binom{8+4-1}{4-1} = \binom{11}{3} = 165$$
   
   Final answer: 165

   417 tokens in, 216 out (104 of them reasoning) · 1.7 s · $0.0002 · 1 message on Pro · answered by deepseek/deepseek-v4.1-flash via Novita · September 27, 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 Pro

   Passed: Final answer 165: right.

   Let the four-digit number be \(abcd\), with \(a \in \{1,\dots,9\}\) and \(b,c,d \in \{0,\dots,9\}\).
   
   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, 218 out (98 of them reasoning) · 3.2 s · $0.0006 · 1 message on Pro · answered by deepseek/deepseek-v4-pro-0813 via Alibaba · September 27, 2026

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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

12. Grok 4.7

    Passed: Final answer 165: right.

    The four-digit positive integers \(ABCD\) (with \(A \in \{1,\dots,9\}\) and \(B,C,D \in \{0,\dots,9\}\)) whose digits sum to 9 are in bijection with the non-negative integer solutions of \(A' + B + C + D = 8\) via the change of variables \(A' = A - 1\).  
    
    No variable can exceed 8, so the upper bounds of 9 are automatically satisfied. The number of solutions is therefore
    \[
    \binom{8+4-1}{3} = \binom{11}{3} = 165.
    \]
    
    Final answer: 165

    1,613 tokens in, 925 out (780 of them reasoning) · 11.1 s · $0.0056 · 1 message on Pro · answered by x-ai/grok-4.7 via xAI · September 27, 2026

13. 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

14. 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

15. 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

16. 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

• Pens at 3 for $4
• A discount, then sales tax
• Compound interest over three years
• The highest of three dice is a 5
• The best AI for math