Thread: Multiples of any natural number - only with 1,0 as digits

I read this somewhere recently.

For any natural number n, there exists a multiple of n, such that the multiple has only 0 and 1 as it's digits.

For e.g for 2, 3, 4, 5, 6 etc we have 10, 111, 100, 10, 1110 etc

Any ideas how to go about proving this?

Originally Posted by aman_cc

I read this somewhere recently.

For any natural number n, there exists a multiple of n, such that the multiple has only 0 and 1 as it's digits.

For e.g for 2, 3, 4, 5, 6 etc we have 10, 111, 100, 10, 1110 etc

Any ideas how to go about proving this?

OEIS

id:A004290 - OEIS Search Results

From OEIS there is a link to this site

Binary

Interesting!

Thanks. Infact I was able to work out the first proof given - that of dividing n into 1, 11, 111, 1111, ,,,,, etc etc

But how to find the smallest such multiple still eludes me. Thanks

Originally Posted by aman_cc

Thanks. Infact I was able to work out the first proof given - that of dividing n into 1, 11, 111, 1111, ,,,,, etc etc

But how to find the smallest such multiple still eludes me. Thanks

If you have time to solve this similar problem

Problem 303 - Project Euler

then in the solution forum you will find some nice related discussion. Unfortunately I don't have time to adapt everything to this problem and then hide the source so people can't cheat.

Problem 303 is pretty easy, but could be hard if you're not already familiar with a programming language.