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.
66336:

The last digit is 6 because 3 is not divisible  by 2.

The second-to-last digit is 3 because 66 is not divisible  by 4.

The third-to-last digit is 3 because 636 is not divisible by 8.

The fourth-to-last digit is 6 because 3336 is not divisible by 16.

The fifth-to-last digit is 6 because 36336 is not divisible by 32.

(Edit: This is sufficient because $2^{n}$ divides $10^{n}$, so it doesn't matter what the preceding digits are.)
Another easy way would be to work with binary representation.

The number in question is a power of 2 and all powers of a number n when represented in the same base n must have their most significant digit as 1 and all other digits as 0.

e.g. 2 in binary is 10, 4 is 100, 8 is 1000...., 1024 is 10,00,00,00,000 and so on. Your number will also look like this.

But, of course, this method is equivalent to sequentially checking divisibility by 2, 4, 8, 16 and 32.