Hi, i didnt know where to put this problem
Show that every positive integer has a multiple made of only sevens and zeros.
thanks for your help.
So we have an integer $\displaystyle n$ and we want an integer $\displaystyle m$ such that $\displaystyle nm = s$, where $\displaystyle s$ consists of only 7's and 0's.
This is a constructive proof, but I'm sure there are other ways of doing it.
Firstly take a number of the form $\displaystyle x = 1111\ldots 111$, constisting only of 1's. Then calculate $\displaystyle r = x \mod{n}$
If $\displaystyle r=0$ then we can simply multiply r by 7 and we are done.
If not then we find 2 other numbers of the form $\displaystyle y_1=111\ldots 111$ > $\displaystyle y_2 = 11\ldots 111$ such that $\displaystyle y_1 \equiv y_2 \mod{n} $. Then $\displaystyle y_1 - y_2 $ is divisible by n and consists only of 1's and 0's, we simply multiply this by 7 and we are done.
The justification for the existence of such $\displaystyle y_i$ is that if we take $\displaystyle n$ elements from the set $\displaystyle \left\{1,11,111,1111,\ldots \right\}$ , then there must be at least 2 with the same remainder mod n.
Pomp.