2021 JMPSC Sprint Problems/Problem 11

Revision as of 19:55, 7 September 2021 by Yofro (talk | contribs) (Solution 2 (General Method))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

See also

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

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Invalid username
Login to AoPS