What is the remainder when 32^32^32 is divided by 7?

Many of you got the correct answer but wrong explanation and it is the biggest losing point of the learning as the same question is never going to come.

trueblue nailed it with correct explanation.

Here is my explanation and I would request the Quant guru of Gmat Club - Bunnel to correct and improve my explanation for the solutions.

\(32^{32^{32}}\) looks quite daunting, but remember one thing on exams like GMAT you will always get tricky and convoluted questions. But they can solved by simple Quant basics.

Always reduce your question.... Rof means remainder of

Rof \(32^{32^{32}}\) when divided by 7 = Rof \(4^{32^{32}}\) by 7 as 28+4 = 32

To calculate Rof \(4^{32^{32}}\) , we need to understand how to reduce it further.

Whenever we are solving remainder questions we always reduce it to the minimum value. To reduce the powers of 4 we need to find \(4^x\) such that when \(4^x\) is divided by 7 the remainder is either 1 or -1

Reason : Rof \(4^{xy}\) = Rof \(4^x * 4^y\) = Rof \(4^{x}\) * Rof \(4^{y}\)

If remainder of Rof \(4^{x}\) =1 , then we can reduce the Rof \(4^{y}\) multiple times until y>x.

Since Rof \(4^{3}\) when divided by 7 is 1, if we can reduce \(4^{32^{32}}\) to the form of \(4^{3k+r}\) we can easily eliminate redundant powers of 4.

To represent \(4^{32^{32}}\) as \(4^{3k+r}\) we need to represent \(32^{32}\) in the form of \(3k+r\).

Now to represent \(32^32\) in the form of \(3k+r\), we have to find the remainder when \(32^{32}\) is divided by 3. That will give the value of r.

Rof \(32^{32}\) when divided by 3 = Rof \(2^{32}\) = Rof \(2^{4*{8}}\)

Since Rof \(2^4\)when divided by 3 = 1, => Rof \(2^{4*{8}}\) = 1

Hence r = 1 => \(32^{32}\) = 3k+1

Now coming back to the main question.

Rof \(4^{32^{32}}\) = Rof \(4^{3k+1}\) = Rof \(4^{3k}\) * Rof \(4^1\) = 1*4 = 4

Hence B.

Whenever you see such question, always apply the above rule.