What is the remainder when 30 raised to the power (7287) is divided by 11?

Concept: When gcd(a, m) = 1 for a modulus m, Fermat's Little Theorem (m prime) guarantees a(m−1) ≡ 1 (mod m). This lets a tower exponent bc be evaluated modulo m by first reducing the huge exponent bc modulo the order of a (a divisor of m − 1), then raising a only to that reduced exponent.

1. Since 11 is prime and gcd(30, 11) = 1, Fermat's Little Theorem gives 3010 ≡ 1 (mod 11), so it is enough to know the exponent 7287 modulo 10.

2. 72 ≡ 2 (mod 10), so reduce 287 modulo 10. The units digits of powers of 2 repeat with period 4: 2, 4, 8, 6, 2, 4, 8, 6, …

3. 87 = 4 × 21 + 3, so 287 lands on the 3rd position of that cycle: 287 ≡ 8 (mod 10). Hence 7287 = 10k + 8 for some integer k.

4. So the original expression, 30 raised to the power (7287), equals 30(10k+8) = (3010)k × 308 ≡ 1k × 308 ≡ 308 (mod 11).

5. 30 ≡ 8 ≡ −3 (mod 11), so 308 ≡ (−3)8 = 38 (mod 11).

6. Powers of 3 mod 11 repeat with period 5, since 35 ≡ 1 (mod 11): 3, 9, 5, 4, 1, 3, 9, 5, 4, 1, … So 38 = 35 × 33 ≡ 1 × 5 = 5 (mod 11).

Cross-check (matching the second route worked in the image): write 30 = 3 × 10. Then 30(10k+8) = 3(10k+8) × 10(10k+8) (mod 11). Since 10 ≡ −1 (mod 11) and 10k + 8 is even, 10(10k+8) ≡ 1. Also (10k + 8) mod 5 = 3 (10k is a multiple of 5), so 3(10k+8) ≡ 33 ≡ 5. Multiplying the two parts gives 5 × 1 = 5 — the same remainder.

So the remainder when 30 raised to the power (7287) is divided by 11 is 5.