CCRT = Constant case CRT: $\,x\equiv a\pmod{\! 2},\ x\equiv a\pmod{\! 5}\iff x\equiv a\pmod{\!10}$ Problem: Find the units digit of $3^{100}$ using Fermat's Little Theorem (FLT).
My Attempt: By FLT we have $$3^1\equiv 1\pmod2\Rightarrow 3^4\equiv1\pmod 2$$ and $$3^4\equiv 1\pmod 5.$$ Since $\gcd(2,5)=1$ we can multiply the moduli and thus, $3^4\equiv 1\pmod {10}\Rightarrow3^{4*25}\equiv 1\pmod{10}.$ So the units digit is $1.$

The phrase ‘Since gcd(2,5)=1 we can multiply the moduli’ is not clear at all. I would rather say something like  ‘since $3^4\equiv 1\mod 2$ and $\bmod5$, we have $3^4\equiv 1\mod \operatorname{lcm}(2,5)=10$’ by the Chinese remainder theorem.
That said, why make things more complicated than they are?
$3^2\equiv -1\mod 10$, hence $3^4\equiv (-1)^2=1\mod 10$, and finally $3^{10}=(3^4)^{25}\equiv 1\mod10$.
