Question

$\forall n\in N,5^{2n}-1$ is divisible by

• A. $6$
• B. $11$
• C. $24$
• D. $26$

Answer 1

Answer: (A), (C)

The correct options are
A $6$
C $24$

Let $P\left(n\right)=5^{2n}-1={25}^n-1=(1+24{)}^n-1$
$\Rightarrow P\left(n\right)=1{+}^n{C}_1.24{+}^n{C}_2{.24}^2+..........{+}^n{C}_n{.24}^n-1$
$\Rightarrow P\left(n\right)=24{(}^n{C}_1{+}^n{C}_2.24+..........{+}^n{C}_n{.24}^{n-1})=24\times$ Integer.
Therefore $P\left(n\right)$ is divisible by $24$ and hence by $6$.