<s>[INST] Need to prove that $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $99$ for all $n \in \mathbb{N} $, using induction. First, obviously, I figured out the base case. So I have $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5 = 99k$ for some $k  \in \mathbb{N} $. As for the inductive step, I was thinking about splitting it up into two parts; proving that the $4\cdot 10^{2n} + 9\cdot 10^{2n-1} + 5$ is divisible by $9$, and that it's divisible by $11$. Unfortunately, I factor it out and get $10(4\cdot10^{2n} + 9\cdot10^{2n-1} + 5)-45$, which is equal to $10(99k)-45$, but of $99$'s divisors, only $3$ divides $45$.
Is there a more ingenious way to factor this equation so that all of my problems are solved?
 [/INST] Assuming that $4\times10^{2n}+9\times10^{2n-1}+5=99k$,
$$
4\times10^{2(n+1)}+9\times10^{2(n+1)-1}+5=4\times10^{2n+2}+9\times10^{2n+1}+5=100\times(4\times10^{2n})+100\times(9\times10^{2n-1})+5=100\times(4\times10^{2n}+9\times10^{2n-1})+5=100\times(4\times10^{2n}+9\times10^{2n-1}+5-5)+5=100\times(4\times10^{2n}+9\times10^{2n-1}+5)-500+5=100\times99k-495=100\times99k-5\times99=(100k-5)\times99.
$$
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