Question

Find the remainder when $7^{103}$ is divided by $25$

Answer 1

Only the last two digits of $7^{103}$ matter, because any number ending in $00$ is divisible by $25$.

If $a\equiv 1\left(mod4\right)$, the last two digits of $7^a$ are $07$

If $a\equiv 2\left(mod4\right)$, the last two digits of $7^a$ are $49$

If $a\equiv 3\left(mod4\right)$, the last two digits of $7^a$ are $43$

If $a\equiv 0\left(mod4\right)$, the last two digits of $7^a$ are $01$

$103\equiv 3\left(mod4\right)$, so the last two digits of $7^a$ are $43$.

$\therefore 43÷25=1$ remainder is $18$