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 $5$ .