If (77 77 +77) is divided by 78, then the remainder will be:

Correct — C. The stem prints the power as "77 77" because the superscript is flattened in extraction; read it as 77⁷⁷ + 77.

Work modulo 78, where the base is one less than the divisor:
77 = 78 − 1 ≡ −1 (mod 78)

77⁷⁷ ≡ (−1)⁷⁷ = −1, because the exponent 77 is odd
the trailing + 77 ≡ −1 as well
sum ≡ −1 + (−1) = −2 (mod 78)

A remainder is reported as a non-negative value, so add the divisor once:
−2 + 78 = 76 → option (c).

Why the others are wrong

• (a)
  78 — 78 is the divisor itself, and a remainder is always smaller than what you divide by — the only values available here are 0 to 77. Working that ends on 78 has really ended on 0.
• (b)
  77 — 77 is the remainder of 77⁷⁷ on its own: −1 ≡ 77 (mod 78). Correct so far, but the stem adds another 77, which is a second −1, carrying the total to −2 ≡ 76.
• (d)
  1 — 1 is what an even exponent would give, since (−1) raised to an even power is +1. The exponent here is 77, an odd number, so the power leaves −1.

This is modular arithmetic, and the whole question turns on one observation: the base 77 sits one step below the divisor 78.

Whenever the base is divisor − 1, its powers alternate. An odd power leaves −1, an even power leaves +1. Nothing has to be expanded, and the size of the exponent is irrelevant beyond its parity.

The second habit worth keeping is that a negative residue is legal in the middle of a calculation but not at the end. Finish by adding the divisor until the value sits between 0 and divisor − 1.

Key facts

• 77 ≡ −1 (mod 78), because 77 = 78 − 1.
• (−1)ⁿ is −1 for odd n and +1 for even n, so 77⁷⁷ ≡ −1 (mod 78).
• A remainder is reported in the range 0 to divisor − 1, so −2 (mod 78) is written as 76.

Common traps

• Stopping at −2 and reporting 2 instead of adding 78 to reach 76.
• Treating the exponent 77 as even because the base is also 77.
• Reading the flattened stem as 77 × 77 + 77, which leaves 0 and matches no option.

No directly related past PYQ was found.