GET your dubs together On 4chan, a popular game is get. Every post on the site gets a sequential post ID. Since you can't influence or determine them, people try to guess (at least a part of) their own post number, usually the first few digits. Another version of the game is called dubs, and it's goal is to get repeating digits at the end of the number (i.e 1234555).
Your task, if you wish to accept it, is to write a program that takes a post id as input (standard integer, you can assume below 2^32), and returns how many repeating digits are on the end.
Rules

Standard loopholes are disallowed.
The program can be a function, full program, REPL command, whatever works, really, as long as no external uncounted code/arguments are needed to run it.
Input can come from STDIN, function arguments, command line argument, file, whatever suits you.

Test Cases
Input: 14892093
Output: 1

Input: 12344444
Output: 5

Input: 112311
Output: 2

Input: 888888
Output: 6

Input: 135866667 //Post number I got on /pol/ few days ago, rip
Output: 1
Mathematica, 29 bytes

How about an arithmetic solution?

IntegerExponent[9#+#~Mod~10]&amp;


I'm very pleased to see that this beats the straight-forward Mathematica approach.

Explanation

The code itself computes 9*n + n%10 and then finds the largest power of 10 that divides the input, or in other words, counts the trailing zeros. We need to show if n ends in k repeated digits, that 9*n + n%10 has k trailing zeros.

Rep-digits are most easily expressed mathematically by dividing a number like 99999 (which is 105-1) by 9 and then multiplying by the repeated digit. So we can write n = m*10k + d*(10k-1)/9, where m ≢ d (mod 10), to ensure that n doesn't end in more than k repeated digits. Note that d = n%10.

Let's plug that into our formula 9*n + n%10. We get 9*m*10k + d*(10k-1) + d. The d at the end is cancelled, so we're left with: 9*m*10k + d*10k = (9*m + d)*10k. But 9 ≡ -1 (mod 10), so 9*m + d ≡ d - m (mod 10). But we've asserted that m ≢ d (mod 10) and hence d - m ≢ 0 (mod 10). 

In other words, we've shown that 9*m + d is not divisible by 10 and therefore, the largest power of 10 that divides 9*n + n%10 = (9*m + d)*10k is k, the number of trailing repeated digits.

As a bonus, this solution prints the correct result, ∞, for input 0.