# Math Help - last two digits

1. ## last two digits

what are the last two digits of $77^{25} * 65^{12}$ ?

2. $\displaystyle (7*11)^{25}=7^{25}*11^{25}$

$\displaystyle 7^5\equiv 07 \ (mod \ 100)$

$\displaystyle 11^{11}\equiv 11 \ (mod \ 100)$

$\displaystyle (7*11)^{25}=7^{25}*11^{25}=(7)^5*(11^{11})^{2+3}\e quiv 7^5*11^5\equiv 7*51\equiv 57 \ (mod \ 100)$

Now work on 65

3. The post above me is not quite correct.

$7^5 \equiv 07 mod 100$

So $7^{25} \equiv (07)^5 \equiv 7 mod 100$.

$11^{11} \equiv 11 mod 100$

So $11^{25} \equiv 11^3*11^{22} \equiv 31*11 \equiv 51 mod 100$.

So, $77^{25} \equiv 7*51 \equiv 57 mod 100$.

So the last two digits of $77^{25}$ are $57$, not $31$.

NOW, work on $65$.

4. I was thinking about inverses and congruence to 1 for some odd reason.

5. Originally Posted by earthboy
what are the last two digits of $77^{25} * 65^{12}$ ?
Taking smaller moduli...

Let $a=77^{25}\cdot65^{12}$. Note that $100=4\cdot25$. We can work separately modulo 25 and then modulo 4.

Clearly $25$ divides $65^{12}$, and thus $25$ divides $77^{25}\cdot65^{12}$. So $a=77^{25}\cdot65^{12}\equiv0\equiv25\pmod{25}$.

Both $77$ and $65$ are congruent to $1$ modulo 4. So $a=77^{25}\cdot65^{12}\equiv1^{25}\cdot1^{12}=1\equ iv25\pmod4$.

We combine $a\equiv25\pmod{25}$ and $a\equiv25\pmod4$ to get $a\equiv25\pmod{100}$. (Since (25,4)=1.)

6. Hello, earthboy!

$\text}What are the last two digits of }77^{25}\cdot65^{12} \,?$

$77^{25}\cdot65^{25} \:=\:(77\cdot65)^{25}$

. . . . . . . . $=\:(5005)^{25}$

. . . . . . . . $\equiv\:(05)^{25} \text{ (mod 100)}$

. . . . . . . . $\equiv\:25\text{ (mod 100)}$