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Difference between revisions of "The perfect squares from $1$ through $2500,$ inclusive, are printed in a sequence of digits $1491625\ldots2500.$ How many digits are in the sequence?"

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So we have a total of <math>1\times3+2\times6+3\times22+4\times19=\boxed{157}</math> digits.
 
So we have a total of <math>1\times3+2\times6+3\times22+4\times19=\boxed{157}</math> digits.
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Latest revision as of 16:53, 14 May 2025

We consider it by four cases:

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

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

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

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

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

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