s(n) is a n-digit number formed by attaching the first n perfect

First two digit number is 10 and first 2 digit square is 16. 16 is the square of 4. This means the first 3 numbers have a square which is a single digit.
First 3 digit number is 100 which is the square of 10. So numbers from 4 till 9 must have 2 digit squares. This gives us a total of 9-4+1 = 6 numbers
First 4 digit number is 1000 and first 4 digit square is 1024 (which is 2^10 or 32^2). So numbers from 10 to 31 must have 3 digit squares. This gives us a total of 31 - 10 + 1 = 22 numbers.
First 5 digit number is 10000 which is the square of 100. So numbers from 32 to 99 must have 4 digit squares. This gives us 99 - 32+1 = 68 numbers

We need to find the total number of digits in all of the square numbers from 1 - 99.

1, 4, 9 <-- all have one digit.
16, 25, 36, ...., 81 <-- all have two digits.

Each 'n' will acquire a new digit when n^2 exceeds a power of ten.

\(sqrt(10)\) is more than 3 and less than 4, so n =1,2,3 will have 1 digit when you square n (1,4,9). This makes 3 numbers.
\(sqrt(100) = 10\), so n = 4,5,6,7,8, and 9 will have two digits when you square n. (16, 25, ...,81). This makes 9 - 3 = 6 numbers.
\(sqrt(1000)\) is more than 30 (30^2 = 900) and a quick check will show that \(31^2 = 961\), while \(32^2 = 1024\), so n = 10,11,...,31 will have 3 digits. This makes 31 - 9 = 22 numbers.
\(sqrt(10000) = 100\), so all of the rest of the numbers from 32-99 will each have 4 digits. This makes 99-31 = 68 numbers.

Therefore, the total number of digits is 3(1) + 6(2) + 22(3) + 68(4) = 3 + 12 + 66 + 272 = 353.

Answer: B.

know the numbers wherein the count increases i.e at 10...all 3 digit squares come....at 31 all 4 digit squares come...till 99
so it is 18 digits till 10, next 21 numbers to go till 32, & lastly 68 numbers till 99...so 18+(21*3)+(68*4)=353

VERITAS PREP OFFICIAL SOLUTION:

Since this exponent-based problem deals with massive numbers, the Guiding Principles of Exponents urge you to look for a pattern. We know that the first three squares (1, 4, 9) have one digit each. We know that the next six squares (16, 25, 36, 49, 64, and 81) have two digits each. Now we need to see when our squares have three digits and four digits, respectively.

We know that 10^2 = 100, which is the smallest number that has three digits. We know that 30^2 = 900, so we're still at three digits. 31^2 = 961, so that works too, but 32^2 will be put us over 1000. Hence the squares from 10 to 31 all have THREE digits.

100^2 = 10,000, which is the smallest number to have five digits, so 99^2 must have four. Hence all the squares from 32 to 99 must have four digits.

Now we just need to add all this up.

1^2 -> 3^2 have one digit each, so 3*1 = 3 total digits

4^2 -> 9^2 have two digits each, so 6*2 = 12 total digits

10^2 -> 31^2 have three digits each, so 22*3 = 66 total digits

32^2 -> 99^2 have four digits each, so 68*4 = 272 total digits

272 + 66 + 12 + 3 = 353, so that's our answer.