What is the remainder when 32^32^32 is divide… | GMAT Sprint

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PS · Problem Solving

705-805

Divisibility/Multiples/Factors

What is the remainder when $32^{32^{32}}$ is divided by 7?

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Official Explanation

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If we use the above approach I'd work with prime as a base.

32^{{32}^{32}}=(28+4)^{{32}^{32}}

. Now if we expand this, all terms but the last one will have 28 as a multiple and thus will be divisible by 7. The last term will be

4^{{32}^{32}}=4^{{(2^5)}^{32}}=4^{2^{160}}=2^{2^{161}}

.

So we should find the remainder when

2^{2^{161}}

is divided by 7.

2^1

divided by 7 yields remainder of 2;

2^2

divided by 7 yields remainder of 4;

2^3

divided by 7 yields remainder of 1;

2^4

divided by 7 yields remainder of 2;…

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What this question tests

Your understanding of divisibility rules, least common multiples, and greatest common factors.

Forgetting that 1 is not prime and that every integer divides 0.