Pure Modular Arithmetic

Note that $n=14^{5}$ .

Now since the base of the number is 7 and also it is prime , we set up a simple congruence :

$7\equiv 1 \mod 2$ , now using the congruence of the type : When $a\equiv b \mod p^{k}$ where $p$ is any prime , then $a^{p}\equiv b^{p} \mod p^{k+1}$ , using this on our congruence , we get , $7^{80}\equiv 1 \mod 2^{5}$ . Now raising the powers of both sides of the congruence , we get , $7^{80 . 9}=7^{720}\equiv 1 \mod 2^{5}$ .

Multiplying the whole congruence by $7^{5}$ we arrive at the final congruence of , $7^{720}\equiv 7^{5} \mod 14^{5}$ And hence the remainder is , $7^{5}=16807$

$\displaystyle\frac{7^{725}}{14^5} = \frac{7^{720}}{2^5} = \frac{(2400+1)^{180}}{32}$ which gives the reminder of 1, however, since we divide the whole equation with $7^5$ , it must multiply back to the fraction to find proper reminder, hence, $7^{725} \equiv 7^5 \pmod{2744\times196}$