Hello Roland.

In this question, you have to see how many digits have a perfect square when the bases are n=1, 2, 3,....., 99.

1^2=1 has 1 digit
2^2=4 has 1 digit
3^2=9 has 1 digit
4^2=16 has two digits
. . .
Then we only have to add all this results.

But, this is a too large list to make. So, we need a simplification.

You can see that:

- if the base is in the set A={1,2,3}: the square has 1 digit.
- if the base is in the set B={4,5,6,7,8,9}: the square has 2 digits.
- if the base is in the set C={10,11,. . . , 31}: the square has 3 digits.
- if the base is in the set D={32,33, . . . , 99}= the square has 4 digits.

Now, we know that in set A there are 3 numbers, in the set B there are 6 numbers, in C there are 22 numbers and in D there are 68 numbers.

So, s(99) has 3*1+6*2+22*3+68*4=353 digits, where the first factor represents the total of numbers in each set and the second factor represents the total of digits that the square has.

So, the correct answer here is B.

I hope this can help you.

I am available if you would like any follow up.