What is the remainder when 3²⁴⁷ is divided by 11?

Since 11 is prime and does not divide 3, Fermat's little theorem gives 3¹⁰ ≡ 1 (mod 11); in fact the powers of 3 cycle with period 5: 3¹=3, 3²=9, 3³=27≡5, 3⁴≡4, 3⁵≡1 (mod 11).

247 = 5 × 49 + 2, so 3²⁴⁷ ≡ 3² = 9 (mod 11).

Why others fail: 3 is 3¹, not 3²; 1 would require the exponent to be a multiple of the actual cycle length 5, but 247 isn't; 5 is 3³.

Tip: The true cycle length of aᵏ mod p can be shorter than p − 1 (Fermat's bound) — check small powers first to find the real period before reducing the exponent.