<s>[INST] Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$ Show that $\displaystyle{\frac{1}{9}(10^n+3 \cdot 4^n + 5)}$ is an integer for all $n \geq 1$
Use proof by induction. I tried for $n=1$ and got $\frac{27}{9}=3$, but if I assume for $n$ and show it for $n+1$, I don't know what method to use. 
 [/INST] No induction necessary.
$$10 + 3\cdot 4^n + 5 \equiv 1 + 3\cdot 4^n + 5 \equiv 6 + 3\cdot 4^n \pmod9$$
Since everything including the base is divisible by three, this reduces to
$$6 + 3\cdot 4^n \pmod9 \iff 2 + 4^n \equiv 2 + 1 \equiv 0 \pmod3$$
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