Question

The remainder when ${32}^{{\left(32\right)}^{\left(32\right)}}$ is divided by 7, is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

The option $d$ is correct.

$Rem{[{32}^{32}}^{32}/7]=Rem[{4^{32}}^{32}/7]$

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 $4^{\left(3k\right)}$, it will leave a remainder of $1$

The number given to us is ${4^{32}}^{3}2$

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 $4^{3k}$

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

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

=> $Rem\left[{4^{32}}^{32}/7\right]=4$