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 $5^{143}$

The first sentence says that whatever power greater than $1$ you pick, $5$ to that power ends in $25$.  Note that the powers of $5$ are $5,25,125,625,3125\ldots $  The end digits are always $25$ after the first one because when you multiply some number of hundreds by $5$ you still have some number of hundreds and when you multiply $25$ by $5$ you get $125$.  The $1$ goes into the hundreds and you are left with $25$.
Then from $3^{40} \equiv 1 \pmod {100}$ you get $3^{280} \equiv 1 \pmod {100}$ by taking the seventh power.  We now want $5^{121} \cdot 3^{32}$ or $5^{143} \cdot 3^{32} \pmod {100}$.  Noting that $5^2 \cdot 3^4 \equiv 1 \pmod {100}$ is new information, it is not derived from the preceding.  Now we can say $(5^2 \cdot 3^4)^8=5^{16} \cdot 3^{32} \equiv 1 \pmod {100}$  We multiply by the remaining factors of $5$, whether $105$ or $127$ and get $25 \pmod {100}$
