Question

The remainder when $2^{47}$ is divided by $7$ is:

Answer 1

Lets observe the cyclicity of 2 when divided by 7.

For $2^1=2,2=2\left(\text{mod}7\right)$

For $2^2=4,4=4\left(\text{mod}7\right)$

For $2^3=8,8=1\left(\text{mod}7\right)$

For $2^4=16,16=2\left(\text{mod}7\right)$

For $2^5=32,32=4\left(\text{mod}7\right)$

After $2,4$ and $1$ the cycle repeats

Therefore, $Rem\left(\frac{2^{47}}{7}\right)=Rem\left(\frac{2^{47\text{mod}3}}{7}\right)=Rem\left(\frac{2^2}{7}\right)=4$