# Thread: Numbers whose decimal digits are 0 and 1

1. ## Numbers whose decimal digits are 0 and 1

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

2. Originally Posted by Bruno J.
Show that for any positive integer $\displaystyle n$, we can find a positive integer $\displaystyle m$ such that the decimal digits of $\displaystyle mn$ are only $\displaystyle 0$'s and $\displaystyle 1$'s.
This is a first impresssion again (not that my last was too succesful) but wouldn't the brute force solution just take $\displaystyle n=a_0+\cdots+a_n10^n$ and let $\displaystyle m=b_0+\cdots+b_m10^m$ multiply the two and then reduce this to a linear algebra question on solving equations?

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

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

4. Originally Posted by simplependulum
Can we just set $\displaystyle m= \frac{1}{n}\left[ \sum_{j=1}^T 10^{b_j \phi(n) } \right ]$

where $\displaystyle T$ is a multiple of $\displaystyle n$ and $\displaystyle b_j , j\in\mathbb{N}$ is any sequence (of course positive ) .
what if $\displaystyle \gcd(n,10) > 1$?

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

6. Originally Posted by simplependulum
Perhaps there are some limits in Euler Phi function but if $\displaystyle (n,10) \neq 1$ then $\displaystyle n$ must be $\displaystyle 2^a 5^b N$ , now $\displaystyle N$ is coprime with $\displaystyle 10$ . I let $\displaystyle m = 5^a 2^b M$ so $\displaystyle mn = 10^{a+b} MN = MN00000000000000000000000...$ .
now it's a complete proof!