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

We'll use the divisiblity 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}

\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^{32 × 32} = 4^{1024}

Number has a specific property,

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

And then the same remainder 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

Here, 4^{1024} = 4^{(3×341+1)}. So, it leaves the remainder 4.