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)

We know that, $5^4=625=52\times 12+1$
$\Rightarrow 5^4=52\lambda +1$ , where $\lambda$ is a positive integer.
$\Rightarrow {\left(5^4\right)}^{24}={\left(52\lambda +1\right)}^{24}$
$=\_{}^{24}{C}_0{\left(52\lambda \right)}^{24}+\_{}^{24}\left(C_1{\left(52\lambda \right)}^{23}+\_{}^{24}{C}_2{\left(52\lambda \right)}^{22}\right)+\dots +\_{}^{24}\left(C_{23}\left(52\lambda \right)+\_{}^{24}{C}_{24}\right)$ (by binomial theorem)
$\Rightarrow 5^{96}=52\left[\_{}^{24}{C}_{0}52^{23}{\lambda }^{24}+\_{}^{24}{C}_{1}52^{23}{\lambda }^{22}+\dots +\_{}^{24}{C}_{23}\lambda \right]+1$
$=\left(amultipleof52\right)+1$
On multiplying both sides by $5$ , we get
$5^{97}=5^{96}\cdot 5=5\left(amultipleof52\right)+5$
Hence, the required remainder is $5$ .