7, 13, and 37 will always divide numbers such as 151515, 474747, 868686 I happened to be looking at two digit numbers that repeated 3 times, such as 151515, 474747, 868686, etc., 3 always goes into these, which is self explanatory because the total of the numbers will always total a number divisible by 3, what threw me is that the numbers 7, 13, and 37 will always go into a number such as this. Why is that?
 A: Forming a number by "repeating a two digit number three times" is the equivalent of "multiplying a two digit number by 10101". And, as @achuille-hui said, $10101 = 3 \times 7 \times 17 \times 37$. So, any number in your form will be a multiple of 3, 7, 17 and 37.
A: The cool thing about 37 and 111 is this: the three identical digits divided by their sum equals 37.  Another fun fact is that 3 × 7 × 37 = 777, which times 13 equals 10,101 (and thus, 151,515 and 474,747)!  The product of 7 × 13 (that is, 91) always goes evenly into multiples of 10,101 because...

*

*91 × 11 = 1,001

*91 × 111 = 10,101

The logic, 111 as a factor of 10,101 (in any base) is not obvious at first glance, but there are two easy ways to prove it is: times 11 = three sets of 11 side-by-side -- 10,101 × 11 = 111,111 -- easily revealing that factor 111!  The alternative: add or subtract a multiple, namely 999, from any three-digit set within, such as switching the middle digit between commas:

*

*111,000 ➡ 101,010

*111,555 ➡ 151,515

*444,777 ➡ 474,747

I just proved (works in every base) 111 is a factor of 10,101! Multiples of 37 (and of 111) are just the coolest, ain't they?
