2710.11 – Sum of Digits

Take $10^{4n^2 + 8} + 1$ . Square it. What is the sum of the digits?

$4$
$4n$
$2 + 2n$
$4 n^2$
$n^2 + n + 2$

Solution

The number $10^{4n^2 + 8} + 1$ looks like $1000\ldots00001$ , that is, it consists of two 1s separated by a lot of 0s. If we square such numbers, we get something like the figure below. So the sum of the digits is 4. The quantity $4n^2 + 8$ is a total red herring. The answer is a).