Find the last two digit of $5^{121}*3^{312}$ The answers key says by using
$$5^k\equiv 25 \pmod{100}, k>=2$$
and $$3^{40}\equiv 1 \pmod{100}$$
can have $$ 5^2 *3^4 \equiv25 \pmod{100},$$
and it follows that the last two digits of $5^{143}*3^{312}$ are $25$.
Don't know why  $ 5^2 *3^4 \equiv25 \pmod{100}$ can applies that the last two digits of $5^{143}*3^{312}$ are $25$.
Also, the question is asking about $5^{121}$, why the last step relates to $5^{143}$

I have no idea what the book is trying to do.
I guess Id do it this way.
$5^2\cdot 3^4 = 25 *81 = 2025 \equiv 25\equiv 5^2 \pmod {100}$.
So $(5^2 \cdot 3^4)^{78} \equiv 25^{78} \pmod {100}$.
SO $5^{156}\cdot 3^{312} \equiv 5^{156} \pmod {100}$.
Now $5^2 \equiv 25 \pmod {100}$ so by induction if $5^k \equiv 25 \pmod {100}$ then $5^{k+1} \equiv 5\cdot 25 =125\equiv 25 \pmod {100}$.
So for any $k, m \ge 2$ you have $5^k \equiv 5^m \equiv 25 \pmod {100}$.
So $5^{156} \equiv 5^{121} \pmod {100}$ and
$5^{121}\cdot 3^{312} \equiv 5^{156}\cdot 3^{312} \equiv 5^{156}\equiv 25 \pmod{100}$.
So the last two digits are $25$.
......
Actually, that's me trying to guess what the book is trying to do.
I'd do:
So $5^{k} \equiv 25$ for all $k \ge 2$.
Then for all $k \ge 2$ then $5^k \cdot 3 \equiv 25 \cdot 3 \equiv 75 \pmod {100}$.
And $5^k 3^2 \equiv 75 \cdot 3 = 225 \equiv 25 \pmod {100}$.
So by induction $5^k 3^m \equiv \begin{cases} 75 & m\ odd\\25 & m\ even\end{cases}$.
So $5^{121} 3^{312} \equiv 25 \pmod {100}$.
