2021 JMPSC Sprint Problems/Problem 11


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


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 $\boxed{48}$.


See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

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