What is the Remainder When \(32^{32^{32}}\) is divided by 7? GMAT Problem Solving

Question: What is the Remainder When \(32^{32^{32}}\) is Divided by 7?

Correct Answer: B

Solution and Explanation
Approach Solution 1:
It is asked in the question that when 32^32^32 is divided with 7, what will be the remainder.

This is a question from Binomial Theorem

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

Now when we apply binomial expansion to this expression then all the terms except the last term will contain the number 28.

28 is divisible by 2 but the last term will be \(4^{32^{32}}\)

Now when this term is divided by 7, it give the required remainder.

\(4^{32^{32}}\) can be rewritten as \(4^{(2^{5})^{32}}\)

= \(4^{2^{160}}\)

Now 4 can be written as\(2^2\)

\((2^2)^{2^{160}}\) = \(2^{{2*2}^{160}}\)=\(2^{2^{161}}\)

The number \(4^{32^{32}}\) is similar to \(2^{2^{161}}\)

Now when\(2^1\) is divided by 7 gives remainder = 2

when \(2^2\) is divided by 7 gives remainder = 4

when \(2^3\) is divided by 7 gives remainder = 1

when \(2^4\) is divided by 7 gives remainder = 2

when \(2^5\) is divided by 7 gives remainder = 4

when \(2^6\) is divided by 7 gives remainder = 1

…

It can be noticed that the remainder has a pattern that repeats itself.
2-4-1 .. 2-4-1..
We have to find \(2^{161}\) comes in which of the three numbers.

\(2^{161}\) is in odd power. 2 in odd power gives 2 according to the pattern.
When 161 will be divided by 2 it will give 5 as remainder.

So the remainder when \(2^{161}\) will be divided by 7 will be same as when\(2^5 \) will be divided by 7

Therefore the remainder will be 4.

Approach Solution 2:
\(32^{32^{32}}\) can also be written as \((2^5)^{{(2^5)}^{2^5}}\)

= \(2^{{5}^{({2^{32*5}})}}\) =\((2^5)^{2^{160}}\) = \(2^{2^{160}*5}\)

= \(2^{5120}\)

\(2^x\) has a cyclicity of 3

Now when\(2^1\) is divided by 7 gives remainder = 2

when\(2^2\) is divided by 7 gives remainder = 4

when \(2^3\) is divided by 7 gives remainder = 1

….

Hence \(2^{5120}\) can be reduced to same as 2^2 / 7 = 4

Approach Solution 3:
Rem[32^32^32/7]=Rem[4^32^32/7]

Now, we need to observe the pattern

4^1 when divided by 7, leaves a remainder of 4

4^2 when divided by 7, leaves a remainder of 2

4^3 when divided by 7, leaves a remainder of 1

And then the same cycle of 4,2, and 1 will continue.

If a number is of the format of 4^(3k+1), it will leave a remainder of 4

If a number is of the format of 4^(3k+2), it will leave a remainder of 2

If a number is of the format of 4^(3k), it will leave a remainder of 1

The number given to us is 4^32^32

Let us find out Rem[Power / Cyclicity] to find out if it 4^(3k+1) or 4^(3k+2). We can just look at it and say that it is not 4^3k

Rem[32^32/3]= Rem[(−1)^32/3]= 1

=> The number is of the format 4^(3k+1)

=> Rem[4^32^32/7]= 4

“What is the Remainder When \(32^{32^{32}}\) is Divided by 7?”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide". To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

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What is the Remainder When \(32^{32^{32}}\) is divided by 7? GMAT Problem Solving

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