Question

If $5^{97}$ is divided by $52$ $,$ then the remainder obtained is

• A. $3$
• B. $5$
• C. $4$
• D. $0$

Answer 1

Answer: (B)

$5$

Answer 2

We know that, $5^{4}=625=52\times 12+1$ $\Rightarrow 5^{4}=52\lambda +1$ , where $\lambda$ is a positive integer. $\Rightarrow \left(5^{4}\right)^{24}=\left(52 \lambda + 1\right)^{24}$ $= \, \\_{}^{24}C_{0}\left(52 \lambda \right)^{24}+\\_{}^{24}\left(C_{1} \left(52 \lambda \right)^{23} + \\_{}^{24}C_{2} \left(52 \lambda \right)^{22}\right)+\ldots +\\_{}^{24}\left(C_{23} \left(52 \lambda \right) + \\_{}^{24}C_{24}\right)$ (by binomial theorem) $\Rightarrow 5^{96}=52\left[\\_{}^{24}C_{0} 52^{23} \lambda ^{24} + \\_{}^{24}C_{1} 52^{23} \lambda ^{22} + \ldots + \\_{}^{24}C_{23} \lambda \right]+1$ $=\left(a \, m u l t i p l e \, o f \, 52\right)+1$ On multiplying both sides by $5$ , we get $5^{97}=5^{96}\cdot 5=5 \, \left(a \, m u l t i p l e \, o f \, 52\right)+5$ Hence, the required remainder is $5$ .