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Difference between revisions of "2024 AMC 10A Problems/Problem 15"
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==Solution==
 
==Solution==
  
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Let <math>M+1213=P^2</math> and <math>M+3773=Q^2</math> for some positive integers <math>P</math> and <math>Q.</math> We subtract the first equation from the second, then apply the difference of squares: <cmath>(Q+P)(Q-P)=2560.</cmath> Note that <math>Q+P</math> and <math>Q-P</math> have the same parity, and <math>Q+P>Q-P.</math>
 +
 +
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
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<cmath>\begin{align*}
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Q+P=1280, \\
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Q-P=2,
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\end{align*}</cmath>
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from which <math>(P,Q)=(639,641).</math>
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 +
~MRENTHUSIASM
  
 
==See also==
 
==See also==
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{{AMC10 box|year=2024|ab=A|num-b=2|num-a=4}}
+
{{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:03, 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 $(P,Q)=(639,641).$

~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

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