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

(Solution)
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The perfect squares are from <math>3^2</math> to <math>50^2</math>. Therefore, the answer is the amount of positive integers between <math>3</math> and <math>50</math>, inclusive. This is just <math>50-3+1=\boxed{48}</math>.
 
The perfect squares are from <math>3^2</math> to <math>50^2</math>. Therefore, the answer is the amount of positive integers between <math>3</math> and <math>50</math>, inclusive. This is just <math>50-3+1=\boxed{48}</math>.
  
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==Solution 2 (General Method)==
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The set is
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<cmath>S=\{3^2,4^2,5^2,\cdots,50^2\}</cmath>
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Notice that the cardinality of
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<cmath>\{a_1,a_2,\cdots,a_n\}</cmath>
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is equal to the cardinality of
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<cmath>\{f(a_1),f(a_2),\cdots, f(a_n)\}</cmath>
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For all functions <math>f</math> with domain containing <math>a_1, \cdots, a_n</math>.
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In our case, apply <math>f(x)=\sqrt{x}</math> to get
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<cmath>S_1=\{3,4,5,\cdots,50\}</cmath>
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Now apply <math>f(x)=x-2</math> to get
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<cmath>S_2=\{1,2,3,\cdots,48\}</cmath>
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Which clearly has cardinality <math>\boxed{50}</math>.
  
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~yofro
  
 
==See also==
 
==See also==

Revision as of 19:54, 7 September 2021

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 $\boxed{50}$.

~yofro

See also

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

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