More Modular in 2016

Setting up a simple congruence which is , $15\equiv 1\mod7$ , and then by using the congruence of type :

If $a\equiv b \mod p^{k}$ where $p$ is any prime number , we have , $a^{p}\equiv b^{p} \mod p^{k+1}$ .

So applying it here , we get : $15^{343}\equiv 1\mod 7^{4}$ , now multiplying whole congruence by $15^2$

we get , $15^{345}\equiv 15^{2}\mod 7^{4}.15^2 = 15^{345}\equiv 225\mod 735^{2}$

And hence remainder is ${\color{#3D99F6} 2}{\color{#3D99F6} 2{\color{#3D99F6} 5}}$

Note : $735^2 = 540225$