## Saturday, July 9, 2011

### Divide Almost Any Odd Number into a Number Consisting of all Nines

Using Discrete Mathematics, we can do so many things. My all-time favorite of them is definitely the amazing proof that states that any odd number that isn't a multiple of five can divide a number consisting of all nines. Sounds hard to believe, but 713 does go into some number consisting of all nines. I'll prove it.

If you think about it, if you divide a number by 713, there are only 713 possible remainders because 713 isn't a remainder. With that in mind, since we are dealing with the numbers consisting of all nines, let's list some out:

9, 99, 999, 9999, ... 9999999...999 (with 714 nines)

Since we have 714 numbers here, two of them must have the same remainder when divided by 713 because there are only 713 possible remainders. Let's take those two numbers, the bigger one is x and the other one is y. Each one of those is equal to 713 times a quotient, two different quotients (say q and p), plus the same remainder (r).

999...999 (with x nines) = 713q + r
999...999 (with y nines) = 713p + r

Technically, we can subtract these two equations from each other. The x nines - the y nines would end up with x - y nines in the front and y zeros at the end because the nines would cancel. According to the Distributive Law, 713q - 713p = 713(q - p) because you can factor a 713 out of both terms. However, the r's at the end cancel each other out, leaving us with no remainder!!! Since q and p are both integers, subtracting them will also be an integer.

999...999 (with x nines) = 713q + r

-  999...999 (with y nines) = 713p + r
999...(x - y nines)...999000...(y zeros)...000 = 713 (q - p)

Since 713 isn't a multiple of 2 or 5, all of the zeros don't matter. Hence, we can cross them all out leaving us with 999...999 (x - y nines) = 713(q - p). Since q - p is an integer, 713 x some integer = a number consisting of all nines. Isn't that cool!!