Remainders of powers repeat in cycles

Correct answer: C

Nobody computes 7^23. Remainders of powers cycle, so find the cycle instead.

Work mod 5. Since 7 ≡ 2 (mod 5), the powers of 7 mod 5 behave like powers of 2:

• 7^1 ≡ 2
• 7^2 ≡ 4
• 7^3 ≡ 8 ≡ 3
• 7^4 ≡ 6 ≡ 1
• 7^5 ≡ 2 again, so the cycle has length 4: 2, 4, 3, 1.

Now place 23 in the cycle. 23 = 4 × 5 + 3, so 7^23 lands on the 3rd position of the cycle, which is 3.

The remainder is 3.

Why the traps exist: (A) would require the power to be a multiple of 5, impossible since 7 and 5 share no factor. (B), (D), and (E) are the other three positions in the cycle, which you get if you misplace the exponent (for example using 23 mod 4 = 3 but reading the cycle one slot off). Write the cycle out with its indices before you match.