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 $1.$

Yours is a valid, clean argument. It is based on this:

If $m$ and $n$ divide $a$, then $lcm(m,n)$ divides $a$.

In your case, you have that $2$ and $5$ divide $3^4-1$, and so $10=lcm(2,5)$ divides $3^4-1$.
