Find the remainder when 23⁵¹ is divided by 100.

Track the last two digits (mod 100) of successive powers of 23: 23¹=23, 23²=529→29, 23³=23×29=667→67, 23⁴=23×67=1541→41, and continuing this way, 23²⁰ ≡ 1 (mod 100) — powers of 23 mod 100 cycle with period 20 (since 23 is coprime to 100, its order divides λ(100)=20). Since 51 = 20×2 + 11, 23⁵¹ ≡ 23¹¹ (mod 100). Computing 23¹¹ mod 100 step by step (squaring and reducing at each stage) gives 23¹¹ ≡ 27 (mod 100). So the remainder is 27. Shortcut: whenever aᵏ ≡ 1 (mod m), reduce the exponent modulo k before evaluating, then use repeated squaring for the residual power.