Last two! #2

We are looking for a two digit number $x$ such that

$x\equiv 29^{46}\mod 100$

Since $100=4\cdot 25$ we can first find $x$ modulo $4$ and $25$ separately

$\begin{aligned} x&\equiv \mathbin{\color{#3D99F6}29}^{46}\\ &\equiv \mathbin{\color{#3D99F6}1}^{46}\equiv 1\mod 4 \end{aligned}$

$\begin{aligned} x&\equiv \mathbin{\color{#D61F06}29}^{46}\\ &\equiv \mathbin{\color{#D61F06}4}^{46\mod \phi (25)}\phantom{ccccccc}\color{#3D99F6}\text{since }\gcd(4,25)=1\text{, we can use Euler's Theorem}\\ &\equiv 4^6\phantom{ccccccccccccccccc}\color{#3D99F6}\phi (25)=20\\ &\equiv 4096\equiv 21\mod 25 \end{aligned}$

We now have the following system of congruences

$\begin{cases} x\equiv 1\mod 4\\ x\equiv 21\mod 25 \end{cases}$

By the Chinese Remainder Theorem, there is only one solution to $x$ modulo $100$ , namely $x\equiv \boxed{21}\mod 100$