what is the remainder when $7^{2015}$ is divided by $25$? The question says that what is the remainder when $7^{2015}$ is divided by $25$. Please help me out, I am out of ideas (I know that I have to find out the last two digits and I will be done... but I only know how to find the units digit.) and I need your help.

I imagine you found the last digit by looking for a pattern when you do powers of 7 like this:
$7^0=1$
$7^1=7$
$7^2=49\equiv9\pmod{10}$
$7^3\equiv9\times7\equiv63\equiv3\pmod{10}$
$7^4\equiv3\times7\equiv21\equiv1\pmod{10}$
So every 4 powers brings you back to 1. So:
$$7^{2015}=7^{4\times503+3}\equiv7^3\equiv3\pmod{10}$$
You can do the same by looking at the remainder each step when you divide by 25.
$7^0=1$
$7^1=7$
$7^2=49\equiv-1\pmod{25}$
$7^3=-1\times7\equiv-7\pmod{25}$
$7^4\equiv-7\times7\equiv-49\equiv1\pmod{25}$
So it follows the same pattern so it has similar working:
$$7^{2015}=7^{4\times503+3}\equiv7^3\equiv-7\equiv18\pmod{25}$$
