Fermat’s theorem and an excluded case

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Work modulo the prime 11.

Part a

Find the remainder when $7^{10}$ is divided by 11, giving a theorem-based justification.

Part b

Find the remainder for $11^{10}$ .

Part c

Show that $7^{9}$ represents a multiplicative inverse of 7 modulo 11.

Model solution and review criteria

Part a

7 is not divisible by 11. Fermat’s little theorem gives $a^{p-1}$ ≡ 1 (mod p), so the remainder is 1.

State primality and coprimality conditions.
Apply the exponent p−1 correctly.

Part b

The base is divisible by 11, so the remainder is 0. The coprimality hypothesis fails.

Handle the excluded case directly.

Part c

Multiply by 7: the product is $7^{10}$ ≡ 1. This establishes an inverse because the base is coprime to the prime.

Use the product congruence.
Retain the required condition.

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