# Find the remainder when \(32^{32^{32}}\) is divided by 7.

We'll use the divisibility rule of 7 and binomial expansion to solve this problem.

## Answer: The remainder when \(32^{32^{32}}\) is divided by 7 is 4.

Let's find the remainder when \(32^{32^{32}}\) is divided by 7.

**Explanation:**

The given number is \(32^{32^{32}}\) = \((28+4)^{32^{32}}\)

If we apply binomial expansion for \((28+4)^{32^{32}}\)

\begin{equation}

(28+4)^{k}={ }^{k} C_{0} 28^{k} 4^{0}+{ }^{k} C_{1} 28^{k-1} 4^{1}+\ldots \ldots \ldots+{ }^{k} C_{k} 28^{0} 4^{k}\end{equation}

If we observe in this expansion, all the terms have 28 as the factor, so all the terms are divisible by 7 except \({ }^{k} C_{k} 28^{0} 4^{k}\)

So, remainder of \((28+4)^{32^{32}}\) = remainder of \((4)^{32^{32}}\)

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

Number 2 has a specific property,

- 2
^{1}divided by 7 gives the remainder as 2 - 2
^{2}divided by 7 gives the remainder as 4 - 2
^{3}divided by 7 gives the remainder as 1 - 2
^{4}divided by 7 gives the remainder as 2 - 2
^{5}divided by 7 gives the remainder as 4 - 2
^{6}divided by 7 gives the remainder as 1

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

Now, we need to know 2^{161} is the 1st, 2nd or 3rd number in the above pattern. 2^{161} is 2 in odd power, 2 in odd power gives a remainder of 2 when divided by cyclicity number 3, so it's the second number in the pattern. This means that the remainder of 2^{161} divided by 7 would be the same as 2^{2} divided by 7. 2^{2} divided by 7 yields a remainder of 4. . So, 2^{161} leaves the remainder of 4.

### Thus, the remainder when \(32^{32^{32}}\) is divided by 7 is 4.

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