Difference between revisions of "2024 AMC 10A Problems/Problem 15"
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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: 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 |
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