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 $5^{143}$
 A: From my point of view, your book may contain some typographical errors. I think what the book really wants to express is as below.
First of all, note that $\forall k \in \mathbb N^* \ \ \ 25^k\equiv 25\mod 100$
Besides, $5^2*3^4\equiv 5^2\mod100$
Note that $5^2$ appears on both sides of the equation, which means that we can use "Recursion" here.
For any $k\in \mathbb N$, $5^2*3^{4*k}=5^2*3^4*3^{4*(k-1)}\equiv 5^2*3^{4*(k-1)}\mod 100$, so keep doing this, we will find that $5^2*3^{4*k}\equiv5^2\mod 100$
Wow, $312=4*78$, so we get $5^2*3^{312}\equiv 25\mod100$
The rest is easy. Just multiply $5^m$($m=143-2$  or  $m=121-2$) on both sides, and you will still get 25 on the right side.
A: $3^{2m}\equiv1\pmod4$
$\implies5^r3^{2m}\equiv5^r\pmod{4\cdot5^r}$
$\equiv5^r\pmod{100}$ for $r\ge2$
Again for $r\ge2,$
$$5^r-25=25(5^{r-2}-1)\equiv0\pmod{100}$$ using Why is $a^n - b^n$ divisible by $a-b$?
A: 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 $25 \pmod {100}$
A: 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 $5^{121} 3^{312} \equiv 25 \pmod {100}$.
A: Clearer: use MDL: $\,\ \color{#c00}ab\bmod \color{#c00}ac\, =\, \color{#c00}a(b\bmod c)\ $ to factor out $\,\color{#c00}{a=5^2}$ from your mod, i.e.
$$\begin{align} 
&\color{#c00}5^{\large \color{#c00}2+k}\,\cdot\,  3^{\large 2n}\ \:\!\bmod 4\cdot \color{#c00}{5^2}\\[.2em]
=\ & \color{#c00}{5^2} (5^{\large k}\cdot  3^{\large 2n}\ \  \bmod 4)\\[.2em]
=\ & 5^2 (1^{\large k} (-1)^{\large 2n}\!\bmod 4)\\[.2em]
=\  & 5^2 (1)
\end{align}\qquad\qquad$$
