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)
$\Rightarrow 5^{96}=13\left[{ }^{24} C_{0} 13^{23} \lambda^{24}+{ }^{24} C_{1} 13^{23} \lambda^{22}\right.$
$\left.+...+{ }^{24} C_{23} \lambda\right]+1$
=(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 $5 .$