The remainder when ( (64)^64)^64 is divided by 7 is equal to: — Solution | JEE Mathematics

MCQEasyJEE 2025Complex Numbers Basics

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

Answer

Correct answer:B

Standard Method

Given: Find the remainder when $\left( (64)^{64} \right)^{64}$ is divided by $7$ .

Find: The correct option.

First reduce the base modulo $7$ .

64 \equiv 1 \mod 7

Now raise both sides to the required powers.

64^{64} \equiv 1^{64} \equiv 1 \mod 7

Therefore,

(64^{64})^{64} \equiv 1^{64} \equiv 1 \mod 7

Thus, the remainder is $1$ . The correct option is B.

Binomial Observation

Given: $64 = 63 + 1$ and the expression is to be divided by $7$ .

Find: The remainder.

Use the observation that any power of $63+1$ leaves remainder $1$ modulo $7$ because $63$ is divisible by $7$ .

(63+1)^{64} = 63\lambda + 1

for some integer $\lambda$ .

Applying the same idea again,

64^{64^{64}} = (63+1)^{64^{64}} = 63\lambda_1 + 1

So the remainder on division by $7$ is $1$ . Hence, the correct option is B.

Reducing the exponent before reducing the base. Here the quickest step is to note that $64 \equiv 1 \mod 7$ . Once the base becomes $1$ , every positive power also remains $1$ modulo $7$ .
Misreading $\left( (64)^{64} \right)^{64}$ as $64^{64} + 64$ or as a product. The whole quantity is a power of a power, so modular reduction must be applied to the base expression, not to a different operation.
Using lengthy cyclicity methods unnecessarily. Although cycles modulo $7$ are valid in general, this question becomes immediate because the base itself leaves remainder $1$ on division by $7$ .

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