We need to find $3^n+ 4^n \mod 5$
Let us take the $3^n \mod 5$ and see the behaviour As 5 is prime we have
$3^4 \equiv 1 \mod 5$
and also
$4^4 \equiv 1 \mod 5$
so $3^n + 4^n \mod 5 $ have period 4
Let us consider $f(n) = 3^n + 4^n \pmod 5$
We have
$f(0) = 2$
$f(1) = 2$
$f(2) = 0$
$f(3) = 1$
And $f(n) = f(4k+n)$
We see that $f(2)$ is zero so $f(4k+2)$ is zero
So we need to find the smallest k suck that $4k + 2 > 99$ and get $k = 25$ and smallest number $102$
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