Remainder when 64^32^32 is divided by 9 is equal to: — Solution | JEE Mathematics

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Remainder when $64^{32^{32}}$ is divided by $9$ is equal to:

Answer

Correct answer:1

Standard Method

Given: Find the remainder when $64^{32^{32}}$ is divided by $9$ .

Find: The numerical remainder.

Use modular arithmetic from the solution:

64 \equiv 1 \pmod{9}

Therefore,

64^{32^{32}} \equiv 1^{32^{32}} \pmod{9}

Since

1^{32^{32}} = 1

the remainder is $1$ . Therefore, the answer is $1$ .

Binomial Expansion Method

Given: Let $t = 32^{32}$ .

Find: The remainder when $64^{32^{32}}$ is divided by $9$ .

Then, as shown in the solution,

64^{32^{32}} = 64^t = 8^{2t} = (9-1)^{2t}

Expanding by the binomial theorem,

(9-1)^{2t} = 9k + 1

for some integer $k$ .

Hence the expression is of the form $9k+1$ , so when divided by $9$ the remainder is $1$ . Therefore, the answer is $1$ .

Taking $64 \equiv 64 \pmod{9}$ without reducing it first is incorrect because remainders must be simplified modulo $9$ . First compute $64 \equiv 1 \pmod{9}$ , then raise to the required power.
Trying to evaluate $32^{32}$ or $64^{32^{32}}$ directly is unnecessary and impractical. Use modular arithmetic instead, because the base already becomes $1$ modulo $9$ .

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