Is there any power of 2 (2^{n}) 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.

#### Solution

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 (2^{2}).

The second to last digit must be 3 since if it were 6, we would have 636 which is not divisible by 8 (2^{3}).

The next digit will be 6 since 3336 is not divisible by 16 (2^{4})

And the fifth digit has to be 6 since 36336 is not divisible by 32 (2^{5})

This is enough since 2^{n} divide to 10^{n} 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 2^{n}.

For the most curious, the first powers of 2 that end with these five digits are:

2^{1196}, 2^{3696}, 2^{6196}... and all solutions have the form 2^{1196 + 2500 * k} for k = 0,1,2, ...