Question

If $n\in N$, then $7^{2n}+2^{3n-3}.3^{n-1}$ is always divisible by

Answer 1

Answer: 25

Putting $n=1$ in $7^{2n}+2^{3n-3}.3^{n-1}$
then, $7^{2\times 1}+2^{3\times 1-3}.3^{1-1}$
$=7^2+2^0.3^0=49+1=\mathrm{50...}(i)$
Also, $n=2$
$7^{2\times 2}+2^{3\times 2-3}.3^{2-1}=2401+24=\mathrm{2425...}(ii)$
From (i) and (ii) it is always divisible by $25$.