JEE Main 2025 Apr 7 Shift 1, Mathematics Q4: Quadratic Equations

JEE Main2025Apr 7, Shift 1Mathematics

The remainder when $((64)^{(64)})^{(64)}$ is divided by 7 is equal to

A
$4$
B
$1$
C
$3$
D
$6$

Solution

Let $N = ((64)^{64})^{64} = 6 4^{6 4^{2}}$ .

Write $64 = 63 + 1 = 7 \cdot 9 + 1$ . So $64 \equiv 1 (mod 7)$ .

Therefore $N = 6 4^{6 4^{2}} \equiv 1^{6 4^{2}} \equiv 1 (mod 7)$ .

Alternatively via the binomial expansion: $N = (1 + 63)^{6 4^{2}} = 1 + (1 6 4 ^{2}) (63) + (2 6 4 ^{2}) (63)^{2} + \dots$ , and every term beyond the first is a multiple of $63 = 7 \cdot 9$ , hence a multiple of 7.

$∴$ Remainder is $1 \Rightarrow$ Option (2).

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