# 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.)