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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.

SacredTxd
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Apr 15 '20 at 07:46
  • Simply specialize the linked divisibility results from polynomials to integers (put $x=10$ there). – Bill Dubuque Apr 15 '20 at 14:33

1 Answers1

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$A(1) = 9$, $A(2) = 9*10+9 = 99$, $A(3) = 9*10^2+9*10+9 = 999$

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 $2$

Aderinsola Joshua
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  • How do the 2nd last line lead to the conclusion? – SacredTxd Apr 15 '20 at 10:00
  • After seeing that $A(2m) = (10^m+1)A(m)$ then $\frac{A(2*m)}{A(m)} = 10^m+1$, if $m=2$ then $A(4)$ is a multiple of $A(2)$, if $m$ was $4$, then $A(8)$ is a multiple of $A(4)$, therefore it is a multiple of $A(2)$ also, so this process continues until we final see that if $m$ is a multiple of $2$, then $A(m)$ is a multiple of $A(2)$ – Aderinsola Joshua Apr 15 '20 at 10:36