The remainder when ((64)⁽⁶⁴⁾)⁽⁶⁴⁾ is divided by 7 is equal to

Solution

Maths sir · 4 steps

Write 64 as 63 + 1, and the huge tower of powers stops mattering.

1. Collapse the tower. A power of a power multiplies the exponents: (64⁶⁴)⁶⁴ = 64^(64 × 64) = 64^N with N = 4096. All that matters is that N is a positive integer.
2. Split the base. 64 = 63 + 1, and 63 = 7 × 9 is a multiple of 7. Expand with the binomial theorem.

   (1 + 63)^N = 1 + C(N, 1)63 + C(N, 2)63² + … + 63^N

3. Spot the multiples. Every term after the first has a factor of 63, so every one of them is divisible by 7. Their sum leaves remainder 0, and only the leading 1 is left over.
4. So. (64⁶⁴)⁶⁴ = 7k + 1 for some integer k, so the remainder on division by 7 is 1, option (d).

The idea to keep. For remainders of big powers, write the base as (multiple of the divisor) ± 1. With +1 the remainder is 1; with −1 it depends on whether the power is even or odd.

Answer from the NTA final answer key. Solution by PrepWiser.