Question

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

• A. 3
• B. 5
• C. 4
• D. 0
• E. 1

Answer 1

Answer: (B)

We know that,
$5^4=625=13\times 48+1$
$\Rightarrow 5^4=13\lambda +1$, where $\lambda$ is a positive integer.
$\Rightarrow {\left(5^4\right)}^{24}=(13\lambda +1{)}^{24}$
$={}^{24}{C}_0(13\lambda {)}^{24}+{}^{24}{C}_1(13\lambda {)}^{23}+{}^{24}{C}_2(13\lambda {)}^{22}$
$+...+{}^{24}{C}_{23}(13\lambda )+{}^{24}{C}_{24}$
(by binomial theorem)
$\Rightarrow 5^{96}=13[{}^{24}{C}_{0}13^{23}{\lambda }^{24}+{}^{24}{C}_{1}13^{23}{\lambda }^{22}$
$+...+{}^{24}{C}_{23}\lambda ]+1$
=(a multiple of 13)+1
On multiplying both sides by 5, we get
$5^{97}=5^{96}\cdot 5=5($ a multiple of 13$)+5$
Hence, the required remainder is $5.$