Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution 2 (General Method))
 
(2 intermediate revisions by 2 users not shown)
Line 4: Line 4:
 
==Solution==
 
==Solution==
 
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==
 +
#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
 +
#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 18:55, 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 $\boxed{48}$.

~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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Sprint_Problems/Problem_11&oldid=161828"