There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number I just took an olympiad and I'm wondering how this problem is solved. 
Problem: There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number. 
Please don't find the solution on your computer and then work backward because you might not be able to fully explain the insight that a person working without a computer would need.

See rogerl's answer to see why you need 32.
Here is a detailed answer.
1) last digit has to be a 6 (duh!)
For last two digits we need either
$$
2^n \equiv 36 \mod 100 \\
2^n \equiv 66 \mod 100
$$
You know that if a number is divisible by 4 and ends in 6, the previous digit has to be odd. So the it has to be 36
Next, one of the following holds
$$
2^n \equiv 336 \mod 1000 \\
2^n \equiv 636 \mod 1000
$$
Since 8 divides 2^n, it has to divide 336 or 636. Clearly only 336 ends.
So the last 3 digits of $2^n$ is 336.
next $X336$ should be divisible by 16. So $X = 6$ so the last four digits are 6336.
Finally $x6336$ should be divisible by 32. So $x=6$
