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

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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>. | ||

+ | ==Solution 2 (General Method)== | ||

+ | The set is | ||

+ | <cmath>S=\{3^2,4^2,5^2,\cdots,50^2\}</cmath> | ||

+ | Notice that the cardinality of | ||

+ | <cmath>\{a_1,a_2,\cdots,a_n\}</cmath> | ||

+ | is equal to the cardinality of | ||

+ | <cmath>\{f(a_1),f(a_2),\cdots, f(a_n)\}</cmath> | ||

+ | For all functions <math>f</math> with domain containing <math>a_1, \cdots, a_n</math>. | ||

+ | In our case, apply <math>f(x)=\sqrt{x}</math> to get | ||

+ | <cmath>S_1=\{3,4,5,\cdots,50\}</cmath> | ||

+ | Now apply <math>f(x)=x-2</math> to get | ||

+ | <cmath>S_2=\{1,2,3,\cdots,48\}</cmath> | ||

+ | Which clearly has cardinality <math>\boxed{48}</math>. | ||

+ | ~yofro | ||

==See also== | ==See also== |

## Latest revision as of 18:55, 7 September 2021

## Problem

How many numbers are in the finite sequence of consecutive perfect squares

## Solution

The perfect squares are from to . Therefore, the answer is the amount of positive integers between and , inclusive. This is just .

## Solution 2 (General Method)

The set is Notice that the cardinality of is equal to the cardinality of For all functions with domain containing . In our case, apply to get Now apply to get Which clearly has cardinality .

~yofro

## See also

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