3

the question is stated as above.

What I have tried:

I have tried to use the remainder theorem to restate the problem. Hence, it would look like this, $\frac {10^m-1}{9}=\frac {k(10^n-1)}{9}$. Clearly, I can get rid of the $9$ and I would have obtained $10^m-1 = k(10^n-1)$. Rearranging the terms would produce

$k.10^n - 10^m = k - 1$ and factorizing it (not sure if it would help); I will get $10^n(k-10^{m-n}) = k-1$.

I am stuck at this line. I was thinking if I were to use Euclid's Lemma to show that $10^n$ does not divide $k-1$, then it must be that $k-10^{m-n}$ divides $k-1$, which would produce $m=n$. However, I clearly cannot prove that $10^n$ does not divide $k-1$ as it is untrue.

Any suggestions how I can prove that the end result is $n|m$? Or should I look at things in another perspective?

Bérénice
  • 9,367
Icycarus
  • 1,142
  • 1
  • 8
  • 21

3 Answers3

5

Theorem : $\forall a, m, n \in \mathbb{Z}^+,\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$.

Proof : See this MSE question

Here $a=10$, so $\gcd(10^n - 1, 10^m - 1) = 10^{\gcd(n, m)} - 1$, so if $10^n-1|10^m-1$, then $\gcd(10^n - 1, 10^m - 1)=10^n-1 $, so $\gcd(n, m)=n$, so $n|m$.

Bérénice
  • 9,367
  • Thanks but there's something I don't get. From the prove, we only can claim that $gcd(10^n-1,10^m-1) = 10^{gcd(n,m)}-1$. How did you manage to prove that if $10^n-1|10^m-1$, then the gcd must be $10^n-1$? Wouldn't the gcd be just $10^{gcd(n,m)}-1$? – Icycarus Apr 22 '17 at 07:06
  • 2
    @Icycarus If $a|b$ then $\text{gcd}(a,b)=a$. In this case, $a=10^n-1$ and $b=10^m-1$. – yurnero Apr 22 '17 at 07:07
  • 1
    @Icycarus Yurnero was faster :), it is a basic property of the $\gcd$. – Bérénice Apr 22 '17 at 07:08
  • @Jennifer Ahh! Yes. I see it now! Thank you! Totally missed that out – Icycarus Apr 22 '17 at 07:10
3

We have this lemma:

If $\frac{x^n-1}{x-1} \bigg| \frac{x^m-1}{x-1}$, for some $x \neq 1$, then $n | m$.

To see this, note that we have nice ways to rewrite these expressions: $$ \frac{x^n-1}{x-1} = \sum_{i=0}^{n-1} x^i $$ and similarly for $m$.

Suppose $n =mq +r$, where $0 \leq r < m$. Then, we collect powers of $m$ from $r$ upto $n$ as follows: $$ x^{r+1},x^{r+2}, \ldots x^{r + m} \\ x^{r+m+1},x^{r+m+2}, \ldots x^{r + m+ m} \\ \ldots \\ x^{r+(q-1)m+1},x^{r+ (q-1)m+1}, \ldots , x^{r+(q-1)m + m} $$ We note the following: $$ x^{r+1},x^{r+2}, \ldots x^{r + m} = x^{r+1}\left(\sum_{i=0}^{m-1} x^i\right)\\ x^{r+m+1},x^{r+m+2}, \ldots x^{r + m+ m} = x^{r+m+1}\left(\sum_{i=0}^{m-1} x^i\right)\\ \ldots \\ x^{r+(q-1)m},x^{r+ (q-1)m+1}, \ldots , x^{r+(q-1)m + m} = r^{r+(q-1)m + 1}\left(\sum_{i=0}^{m-1} x^i\right) $$

The up shot of the above, is that we can write: $$ \sum_{i=0}^n x^i = \left(\sum_{i=0}^{m-1} x^i\right) \left(x^{r+1} + x^{r+m+1} + \ldots + x^{r + (q-1)m + 1}\right) + \sum_{i=0}^r x^i $$

Now remind yourself of the polynomial remainder theorem. Since the degree of the remainder here is $r$, which is less than $m$, we have by uniqueness of the remainder that $\sum_{i=0}^r x^i = 0$, which gives that $r=0$. Hence $m|n$.

Now put $x = 10$ for your theorem.

Sarvesh Ravichandran Iyer
  • 76,105
0

To say $\frac{10^n-1}9\mid\frac{10^m-1}9$ is the same as $(10^n-1)\mid(10^m-1)$. This is the same as $10^m\equiv1\pmod{10^n-1}$. Modulo $10^n-1$ the sequence of powers of $10$ (starting with $10^0$) is $$1,10,100,\ldots,10^{n-2},10^{n-1},1,10,100,\ldots,10^{n-2},10^{n-1},1,10,100,\ldots. $$

Angina Seng
  • 158,341