# Let $N_b = 1_b + 2_b + \cdots + 100_b$ . Find the number of values of $b$ for which the sum of the squares of the digits of $N_b\leq512$.

The problem is:

"Let $b$ be an integer greater than 2, and let $N_b = 1_b + 2_b + \cdots + 100_b$ (the sum contains all valid base $b$ numbers up to $100_b$). Compute the number of values of $b$ for which the sum of the squares of the base $b$ digits of $N_b$ is at most 512."

I could only perform the following step in an attempt to solve the problem: $N_b=1_b+2_b+\cdots+100_b=1+2+\cdots+b^2=\frac{(b^2)(b^2+1)}{2}$

I am not able to figure out what to do next. I have tried using small examples, tried to divide the problem into two cases:b is even and b is odd, and tried to see if there's any polynomial I could write to find out $\frac{(b^2)(b^2+1)}{2}$'s digits. These did not help me in proceeding any further. I would like to have some hint for an appropriate approach to the problem. Any help is appreciated.

HINT: $b^2=100_b$ and $b^2+1=101_b$. Therefore:
• If $b$ is even, then $b^2$ is even, so $N_b=\frac{100_b}{2}\cdot 101_b$. To find the digits of $\frac{100_b}{2}$, use the fact that $\frac{10_b}{2}=\left(\frac b 2\right)_b$.
• If $b$ is odd, then $b^2+1$ is even, so $N_b=\frac{101_b}{2}\cdot 100_b$. To find the digits of $\frac{101_b}{2}$, use the fact that $101_b=(b-1)(b-1)_b+2$.