Question

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

Answer 1

Step 1: Expand given number

The given number is ${\left({32}^{32}\right)}^{32}={\left[{\left(28+4\right)}^{32}\right]}^{32}$

Let us consider

${\left({32}^{32}\right)}^{32}={\left(32\right)}^k={\left(28+4\right)}^k$

Using binomial expansion we get,

${\left(28+4\right)}^k={}^{k}C_0{28}^k{4}^0+{}^{k}C_1{28}^{k-1}{4}^1+........+{}^{k}C_k{28}^0{4}^k$

The expansion has all the factors divisible by $7$ except ${}^{k}C_k{28}^0{4}^k$.
Step 2: Determine the cyclicity

Hence

$4^1$ 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 process of $4,2,$ and $1$ will continue.

If the given number is of form $4^{\left(3k+1\right)}$, a remainder of $4$ is obtained.

If the given number is of form $4^{\left(3k+2\right)}$, a remainder of $2$ is obtained.

If the given number is of form $4^{\left(3k+1\right)}$, it leaves a remainder of $1$.

Step 3: Calculate the remainder

The number given is ${\left(4^{32}\right)}^{32}$

Let us find out $Rem\left[\frac{Power}{Cyclicity}\right]$.

$Rem\left(\frac{{32}^{32}}{3}\right)=Rem\left[\frac{{\left(-1\right)}^{32}}{3}\right]=1$

The number is of form $4^{\left(3k+1\right)}$.

$\therefore Rem\left[\frac{{\left(4^{32}\right)}^{32}}{7}\right]=4$

Hence, the remainder when ${\left({32}^{32}\right)}^{32}$ is divided by $7$ is $4$.