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 "2024 AMC 10A Problems/Problem 15"
Line 10: Line 10:
 
We wish to maximize both <math>P</math> and <math>Q,</math> so we maximize <math>Q+P</math> and minimize <math>Q-P.</math> It follows that
 
We wish to maximize both <math>P</math> and <math>Q,</math> so we maximize <math>Q+P</math> and minimize <math>Q-P.</math> It follows that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
Q+P=1280, \\
+
Q+P&=1280, \\
−
Q-P=2,
+
Q-P&=2,
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 
from which <math>(P,Q)=(639,641).</math>
 
from which <math>(P,Q)=(639,641).</math>
 +
 +
Note that <math>M=P^2-1213=Q^2=3773\equiv1-3\equiv8\pmod{10},</math> so the units digit of <math>M</math> is <math>\boxed{\textbf{(E) }8}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 16:06, 8 November 2024

Problem

Let $M$ be the greatest integer such that both $M+1213$ and $M+3773$ are perfect squares. What is the units digit of $M$?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8$

Solution

Let $M+1213=P^2$ and $M+3773=Q^2$ for some positive integers $P$ and $Q.$ We subtract the first equation from the second, then apply the difference of squares: \[(Q+P)(Q-P)=2560.\] Note that $Q+P$ and $Q-P$ have the same parity, and $Q+P>Q-P.$

We wish to maximize both $P$ and $Q,$ so we maximize $Q+P$ and minimize $Q-P.$ It follows that \begin{align*} Q+P&=1280, \\ Q-P&=2, \end{align*} from which $(P,Q)=(639,641).$

Note that $M=P^2-1213=Q^2=3773\equiv1-3\equiv8\pmod{10},$ so the units digit of $M$ is $\boxed{\textbf{(E) }8}.$

~MRENTHUSIASM

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_15&oldid=231856"