Question

The remainder when $2^{120}$ is divided by $7$.

• A. $3$
• B. $5$
• C. $6$
• D. $1$

Answer 1

The correct option is D $1$

$2^{120}=(2^3{)}^{40}=8^{40}=(7+1{)}^{40}$
So, by binomial expansion we can expand it as
$(7+1{)}^{40}={}^{40}{C}_0{7}^{40}{1}^0+{}^{40}{C}_1{7}^{39}{1}^1+...+..+{}^{40}{C}_{40}{7}^0{1}^{40}$
This can be written as $7m+1$, where $m={}^{40}{C}_0{7}^{39}{1}^0+{}^{40}{C}_1{7}^{38}{1}^1+...+{}^{40}{C}_{39}{7}^0{1}^{39}$
So, the remainder is $1$