Numbers whose decimal digits are 0 and 1

**Bruno J.** · May 7th 2010, 11:35 AM

Show that for any positive integer $n$ , we can find a positive integer $m$ such that the decimal digits of $mn$ are only $0$ 's and $1$ 's.

**Drexel28** · May 7th 2010, 10:10 PM

Originally Posted by Bruno J.

Show that for any positive integer $n$ , we can find a positive integer $m$ such that the decimal digits of $mn$ are only $0$ 's and $1$ 's.

This is a first impresssion again (not that my last was too succesful) but wouldn't the brute force solution just take $n=a_0+\cdots+a_n10^n$ and let $m=b_0+\cdots+b_m10^m$ multiply the two and then reduce this to a linear algebra question on solving equations?

**simplependulum** · May 7th 2010, 10:33 PM

Can we just set $m= \frac{1}{n}\left[ \sum_{j=1}^T 10^{b_j \phi(n) } \right ]$

where $T$ is a multiple of $n$ and $b_j , j\in\mathbb{N}$ is any sequence (of course positive ) .

**NonCommAlg** · May 7th 2010, 11:08 PM

Originally Posted by simplependulum

Can we just set $m= \frac{1}{n}\left[ \sum_{j=1}^T 10^{b_j \phi(n) } \right ]$

where $T$ is a multiple of $n$ and $b_j , j\in\mathbb{N}$ is any sequence (of course positive ) .

what if $\gcd(n,10) > 1$ ?

**simplependulum** · May 7th 2010, 11:22 PM

Originally Posted by NonCommAlg

what if $\gcd(n,10) > 1$ ?

Perhaps there are some limits in Euler Phi function but if $(n,10) \neq 1$ then $n$ must be $2^a 5^b N$ , now $N$ is coprime with $10$ . I let $m = 5^a 2^b M$ so $mn = 10^{a+b} MN = MN00000000000000000000000...$ .

**NonCommAlg** · May 10th 2010, 10:37 AM

Originally Posted by simplependulum

Perhaps there are some limits in Euler Phi function but if $(n,10) \neq 1$ then $n$ must be $2^a 5^b N$ , now $N$ is coprime with $10$ . I let $m = 5^a 2^b M$ so $mn = 10^{a+b} MN = MN00000000000000000000000...$ .

now it's a complete proof!

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