# Difference between revisions of "2021 JMPSC Sprint Problems/Problem 11"

## Problem

How many numbers are in the finite sequence of consecutive perfect squares $$9, 16, 25, \ldots , 2500?$$

## Solution

The perfect squares are from $3^2$ to $50^2$. Therefore, the answer is the amount of positive integers between $3$ and $50$, inclusive. This is just $50-3+1=\boxed{48}$.

## Solution 2 (General Method)

The set is $$S=\{3^2,4^2,5^2,\cdots,50^2\}$$ Notice that the cardinality of $$\{a_1,a_2,\cdots,a_n\}$$ is equal to the cardinality of $$\{f(a_1),f(a_2),\cdots, f(a_n)\}$$ For all functions $f$ with domain containing $a_1, \cdots, a_n$. In our case, apply $f(x)=\sqrt{x}$ to get $$S_1=\{3,4,5,\cdots,50\}$$ Now apply $f(x)=x-2$ to get $$S_2=\{1,2,3,\cdots,48\}$$ Which clearly has cardinality $\boxed{48}$.

~yofro