<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] A number is a multiple of $9$ iff the sum of its decimal digits is a multiple of $9$. Your numbers satisfy this: $4+9+5=18$.
A number is a multiple of $11$ iff the alternating sum of its decimal digits is a multiple of $11$. Your numbers satify this: $4-9+5=0$.
Explicitly, we have $4900\cdots05=9\cdot 544\cdots45$ and $4900\cdots05=11\cdot 4454\cdots5455$.
Or directly, $4900\cdots05=99\cdot 4949\cdots495$, which is the formula given by Martin.
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