# Determination of the last two digits of $777^{777}$

May I know if my proof is correct? Thank you.

This is equivalent to finding $x$ such that $777^{777} \equiv x \pmod{100}.$

By Euler’s theorem, $777^{\ \psi(100)} =777^{\ 40}\equiv 1 \pmod{100}$.

It follows that $777^{760} \equiv 1 \pmod{100}$ and $777^{\ 17} \equiv x \pmod{100}.$

By Binomial expansion, $777^{\ 17} = 77^{\ 17}+700m$, for some positive integer $m$.

Hence $77^{17} \equiv x \pmod{100} \Longleftrightarrow \ x= 97$.

#### Solutions Collecting From Web of "Determination of the last two digits of $777^{777}$"

Yep! Your proof is correct. I particularly like your use of the binomial expansion to make the terms easier to deal with.

Another way : As $\displaystyle777\equiv77\pmod{100}$

and as Carmichael function $\displaystyle\lambda(100)=20$ with $777\equiv17\pmod{20}$ and $\displaystyle(777,100)=1$

$\displaystyle777^{777}\equiv77^{17}\pmod{100}$

Now using Binomial Theorem $\displaystyle77^{17}=(70+7)^{17}\equiv7^{17}+\binom{17}17^{16}\cdot70\pmod{100}$

Again observe that $7^2=49=50-1$
$\displaystyle\implies 7^4=(50-1)^2=50^2-2\cdot50\cdot1+1^2\equiv1\pmod{100}$

As $\displaystyle4|16$ and $\displaystyle17\equiv1\pmod4; 7^{16}\equiv1\pmod{100},7^{17}\equiv7\pmod{100}$

I leave the rest for you complete