2024 AMC 10A Problems/Problem 15: Difference between revisions
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==Solution== | ==Solution== | ||
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 | |||
<cmath>\begin{align*} | |||
Q+P=1280, \\ | |||
Q-P=2, | |||
\end{align*}</cmath> | |||
from which <math>(P,Q)=(639,641).</math> | |||
~MRENTHUSIASM | |||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b= | {{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 
be the greatest integer such that both 
and 
are perfect squares. What is the units digit of ?
Solution
Let 
and 
for some positive integers 
and 
We subtract the first equation from the second, then apply the difference of squares: ![\[(Q+P)(Q-P)=2560.\]](http://latex.artofproblemsolving.com/8/0/8/808b886dfcc0c467f633a6e01b1d195b732cd4df.png)
Note that 
and 
have the same parity, and 
We wish to maximize both and 
so we maximize and minimize 
It follows that
from which 
~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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
