2015
Problem - 2611
What is the $22^{nd}$ positive integer $n$ such that $22^n$ ends in a $2$?
This problem can be solved by looking for patterns.
The last digit of the powers of $22^n$ cycles $2, 4, 8, 6, 2, 4, 8, \cdots$. Therefore, the answer is the $22^{nd}$ term of this arithmetic sequence which is $\boxed{85}$.