View source for The perfect squares from $1$ through $2500,$ inclusive, are printed in a sequence of digits $1491625\ldots2500.$ How many digits are in the sequence?

We consider it by four cases:

<math>\bullet</math> Case 1: There are <math>3</math> perfect squares that only have <math>1</math> digit, <math>1^{2},</math> <math>2^{2},</math> and <math>3^{2}.</math>

<math>\bullet</math> Case 2: The smallest perfect square that has <math>2</math> digits is <math>4^{2},</math> and the largest is <math>9^{2},</math> so that's a total of <math>6</math> perfect squares with <math>2</math> digits.

<math>\bullet</math> Case 3: The smallest perfect square with <math>3</math> digits is <math>10^{2},</math> and the largest is <math>31^{2},</math> yielding a total of <math>22.</math>

<math>\bullet</math> Case 4: The smallest perfect square with <math>4</math> digits is <math>32^{2},</math> and the last one that is no greater than <math>2500</math> is <math>50^{2},</math> giving a total of <math>19.</math>

So we have a total of <math>1\times3+2\times6+3\times22+4\times19=\boxed{157}</math> digits.