Modular Arithmetic ( using Euler's Totient Theorem )

Question

Please help me find the last two digits of $312^{84}$ ; I am stuck right here: $312^\phi \equiv 1 \pmod{100}$.

Refer to the megathread: https://math.stackexchange.com/questions/81228/how-do-i-compute-ab-bmod-c-by-hand?noredirect=1&lq=1 — player3236, Nov 15 '20 at 22:45
Welcome to Mathematics Stack Exchange. Do you see how Euler's theorem helps? Where did you get stuck? Do you know the Chinese remainder theorem? — J. W. Tanner, Nov 15 '20 at 22:52
In case it's not clear, you need to compute $312^{84} \bmod 100$. That already allows some simplification. — lhf, Nov 15 '20 at 22:57

score 3 · Accepted Answer · answered Nov 15 '20 at 23:03

$\phi(100)=40$, but you can't say by Euler's theorem that $312^{40} \equiv1\bmod100$,

because $\gcd(312,100)\ne1$.

Rather, use Euler's theorem modulo $25$ in conjuction with the Chinese remainder theorem.

$312^{84}\equiv12^{84}\equiv12^4\equiv144^2\equiv(-6)^2\equiv36 \bmod25$,

since $12^{20}\equiv1\bmod25$ by Euler's theorem.

$312^{84}\equiv0\bmod4$, so $312^{84}\equiv36\bmod100,$

so the last two digits of $312^{84}$ are $36$.

1 Answers1