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Run on October 15, 2026: 5 inputs on 20 models, each with this tool's own instructions.

Solve this math problem step by step. Number the steps, check the result, and end with a line that starts "Final answer:".

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What is the remainder when 7^100 is divided by 13?

1. Reduce powers of 7 modulo 13:
   7^2 = 49 ≡ 10 (mod 13).
   7^3 ≡ 10 × 7 = 70 ≡ 5 (mod 13).
   7^6 ≡ 5^2 = 25 ≡ 12 ≡ -1 (mod 13).

2. Therefore, 7^12 ≡ (-1)^2 = 1 (mod 13).

3. Write 100 = 8 × 12 + 4, so:
   7^100 ≡ (7^12)^8 × 7^4 ≡ 7^4 (mod 13).

4. Compute the remaining power:
   7^4 ≡ (7^2)^2 ≡ 10^2 = 100 ≡ 9 (mod 13).

5. Check: 100 = 13 × 7 + 9, so the remainder is indeed 9.

Final answer: 9

Step 1: Find the pattern of powers of 7 modulo 13.
7^1 = 7 (mod 13), so 7^1 ≡ 7.

Step 2: Compute the next powers.
7^2 = 49. Since 49 = 3·13 + 10, 7^2 ≡ 10 (mod 13).

Step 3: Continue.
7^3 ≡ 10·7 = 70 = 5·13 + 5, so 7^3 ≡ 5 (mod 13).

Step 4: Continue.
7^4 ≡ 5·7 = 35 = 2·13 + 9, so 7^4 ≡ 9 (mod 13).

Step 5: Continue.
7^5 ≡ 9·7 = 63 = 4·13 + 11, so 7^5 ≡ 11 (mod 13).

Step 6: Continue.
7^6 ≡ 11·7 = 77 = 5·13 + 12, so 7^6 ≡ 12 (mod 13).

Step 7: Notice 12 ≡ -1 (mod 13), so 7^6 ≡ -1 and 7^12 ≡ (-1)^2 = 1 (mod 13).
The order of 7 modulo 13 divides 12, and we have 7^6 ≡ -1 ≠ 1, so the order is 12 (we also checked the smaller candidates 1, 2, 3, 4, 6 are not valid since 7^6 ≠ 1 and 7^4 ≠ 1, 7^3 ≠ 1, 7^2 ≠ 1, and the order must divide 12 and cannot be 6, so it is 4 or 12 or 3 or 2 or 1; 7^4 ≡ 9 ≠ 1 rules out 1, 2, 4, and 7^3 ≡ 5 ≠ 1 rules out 3, so the order is 12).

Step 8: Reduce the exponent.
100 = 8·12 + 4, so 7^100 = (7^12)^8 · 7^4 ≡ 1^8 · 7^4 ≡ 7^4 (mod 13).

Step 9: Use the value from Step 4.
7^4 ≡ 9 (mod 13).

Step 10: Check.
Direct check by another route: 7^4 = 2401. Dividing 2401 by 13: 13·184 = 2392, and 2401 − 2392 = 9. This matches 9.
Also check 7^100 ≡ 7^4 via Fermat's little theorem: 7^12 ≡ 1 (mod 13), and 100 mod 12 = 4, which agrees with Step 8.

Final answer: 9

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