Is there any power of 2 (2n) such that the last five digits of the result is a combination of three or six? Is about find the last five digits of that value.
For example, ... 36363 would not be a valid solution and no power of two would return a result ending in 3.
To find the answer, we can follow the following deduction:
The last digit must be 6 since if it were 3, the number would not be divisible by 2.
The penultimate digit must be 3 since if it were 6, we would have 66 that is not divisible by 4 (22).
The second to last digit must be 3 since if it were 6, we would have 636 which is not divisible by 8 (23).
The next digit will be 6 since 3336 is not divisible by 16 (24)
And the fifth digit has to be 6 since 36336 is not divisible by 32 (25)
This is enough since 2n divide to 10n so no matter what the preceding digits are, all powers of 2 whose last 5 digits of the result are a combination of three and six, will end in 66336.
This is the only possibility since in base 10, the divisibility of the last n digits determines the divisibility between 2n.
For the most curious, the first powers of 2 that end with these five digits are:
21196, 23696, 26196... and all solutions have the form 21196 + 2500 * k for k = 0,1,2, ...