Let A(n) be a repdigit containing n copies of 9. E.g. A(2) = 99.
Prove that n must be even for A(2) to divide A(n).
As such, if Bn is n copies of x. For what values of n will B(2) divide B(n)?
I have solved the problem by showing that 11|A(n) for even n (and clearly 9|A(n) for all n). However, I would like an alternative solution, as well as to generalise the solution for other repdigits.
$A(1) = 9$, $A(2) = 9*10+9 = 99$, $A(3) = 9*10^2+9*10+9 = 999$
Therefore $A(n) = 9*10^{n-1} + A(n-1)$
$A(n) = 10^n-1$
Say $n=2*m$
$A(2*m) = 10^{2*m}-1$
Now factor it $A(2*m) = (10^m+1)*(10^m-1)$
But $A(m) = 10^m-1$
$A(2*m) = (10^m+1)*A(m)$
Therefore $n$ must be a multiple of $2$ for $A(2)$ to divide $A(n)$, $n$ is just just even but a multiple of $2$