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Let $N_n$ be an integer formed of $n$ consecutive $1$s. For example $N_3 = 111,$ $N_7 = 1 111 111.$ Show that $N_n \mid N_m$ if and only if $n \mid m.$

Cameron Buie
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David
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2 Answers2

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Hint: Show that $\,f_n\,$ satisfies $\, f_n \equiv f_{n-m}\pmod{f_m}\,$ for $\,m > n,\,$ hence, by mimicing the Euclidean algorithm, one easily deduces that $\,\gcd(f_m,f_n) = f_{\gcd(m,n)}.\,$ See this answer for more.

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Key Ideas
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Note that $$N_n=\frac{10^n-1}9$$

Bill Dubuque once proved the following

Let $\{f_n\}$ be a sequence of integers with $f_0=0$ such that, when $m>n$; we have $$f_n\equiv f_{n-m}\mod f_m$$ Then $(f_n,f_m)=f_{(n,m)}$

Apply this to any sequence of the form $$\hat a_n=\frac{a^n-1}{a-1}$$ to prove the claim $\hat a_n\mid \hat a_m\iff n\mid m$. In your case $a=10$.

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Pedro
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  • Steven, since you're new, you may not be aware of this: when you find an answer to be particularly helpful, we encourage you to accept it. (But you can only accept one answer per question.) To accept an answer, just click on the $\large \checkmark$ to the left of the answer you'd like to accept! – amWhy Jun 07 '13 at 00:52
  • Nice, Peter! $+1$ – amWhy Jun 07 '13 at 00:53
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    +1 You don't need something that strong. You can show that $\gcd( a^n -1, a^m -1 ) = a^{\gcd(n,m)} - 1 $, which gives the result immediately. – Calvin Lin Jun 07 '13 at 00:58
  • @CalvinLin Sure, but what you give is a special case of this, and the proof of the above is not hard. Also, I wanted to link to one of Bill's many nice answers. – Pedro Jun 07 '13 at 01:03
  • Yea, i know it's a special case (and both have similar proofs). However, the general case is often rarely used, which this special case is often used. – Calvin Lin Jun 07 '13 at 01:05
  • how do you show that Calvin? – David Jun 07 '13 at 01:10