## Online Course Discussion Forum

### Math Challenge II-A Spring 2022 Q10.10

Math Challenge II-A Spring 2022 Q10.10

Hello,

We didn't get through the last question, so I was wondering if the final answer is 1?

Since 11^2011 = 1(mod 8)

and 11^2011 = 121(mod 125)

The Chinese Remainder Theorem can manipulate them to 121(mod 1000) which would leave 1 in the hundreds place.

Is this correct?

Thank you,

Mandy C

Re: Math Challenge II-A Spring 2022 Q10.10

I think you need to be a little careful here with the calculations.

For (mod 8), $11\equiv 3\pmod{8}$, so aren't all the even powers $\equiv 1\pmod{8}$, not the odd ones?

As a hint for the (mod 125) portion. By Euler's Totient Function ($\phi(125) = 100$), $11^{100}\equiv 1 \pmod{125}$ so we are left with $11^{11}\pmod{125}$. This isn't too bad to "bash" out. Hint: $11^2\equiv -4 \pmod{125}$.

Then as you mentioned, you need to put these two together to find the answer (mod 1000).

To see the spoiled answer (remember we just want the hundreds digit) click here for the calculation on Wolfram Alpha.