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 Invitationals Problems/Problem 11"

(→‎Solution)
Line 14: Line 14:
  
 
Therefore the arithmetic progression must not reach <math>8464</math>.  This means the desired answer is <math>\boxed{8459}.</math> ~djmathman
 
Therefore the arithmetic progression must not reach <math>8464</math>.  This means the desired answer is <math>\boxed{8459}.</math> ~djmathman
 +
 +
==See also==
 +
#[[2021 JMPSC Invitational Problems|Other 2021 JMPSC Invitational Problems]]
 +
#[[2021 JMPSC Invitational Answer Key|2021 JMPSC Invitational Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Revision as of 16:29, 11 July 2021

Problem

For some $n$, the arithmetic progression \[4,9,14,\ldots,n\] has exactly $36$ perfect squares. Find the maximum possible value of $n.$

Solution

First note that the integers in the given arithmetic progression are precisely the integers which leave a remainder of $4$ when divided by $5$.


Suppose a perfect square $m^2$ is in this arithmetic progression. Observe that the remainders when $0^2$, $1^2$, $2^2$, $3^2$, and $4^2$ are divided by $5$ are $0$, $1$, $4$, $4$, and $1$, respectively. Furthermore, for any integer $m$, \[(m+5)^2 = m^2 + 10m + 25 = m^2 + 5(2m + 5),\] and so $(m+5)^2$ and $m^2$ leave the same remainder when divided by $5$. It follows that the perfect squares in this arithmetic progression are exactly the numbers of the form $(5k+2)^2$ and $(5k+3)^2$, respectively.


Finally, the sequence of such squares is \[(5\cdot 0 + 2)^2, (5\cdot 0 + 3)^2, (5\cdot 1 + 2)^2, (5\cdot 1 + 3)^2,\cdots.\]

In particular, the first and second such squares are associated with $k=1$, the third and fourth are associated with $k=2$, and so on. It follows that the $37^{\text{th}}$ such number, which is associated with $k=18$, is \[(5\cdot 18 + 2)^2 = 92^2 = 9409.\]

Therefore the arithmetic progression must not reach $8464$. This means the desired answer is $\boxed{8459}.$ ~djmathman

See also

  1. Other 2021 JMPSC Invitational Problems
  2. 2021 JMPSC Invitational 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_Invitationals_Problems/Problem_11&oldid=158092"