When (7777 + 77) is divided by 78, the remainder is:

Concept: For positive integers x and y and an odd exponent n, the sum xn + yn always factors as (x + y) times an integer -- so xn + yn is exactly divisible by (x + y) whenever n is odd. This identity turns a huge power-of-a-sum expression into an easy divisibility check against (x + y), without ever expanding the power.

1. Split the given expression so one part matches the xn + yn pattern: 7777 + 77 = (7777 + 177) + 76, using 77 = 1 + 76.

2. Match x = 77, y = 1, n = 77 (odd) to the identity: 7777 + 177 is divisible by (77 + 1) = 78.

3. So 7777 + 77 = 78k + 76 for some integer k -- the entire divisible block contributes nothing to the remainder.

4. Since 0 ≤ 76 < 78, the remainder on dividing by 78 is exactly 76.

Cross-check: Independently, working modulo 78: 77 = -1 (mod 78), so 7777 = (-1)77 = -1 (mod 78). Adding the extra 77 gives 7777 + 77 = -1 + 77 = 76 (mod 78), confirming the same remainder.

Result: The remainder is 76.