Difference between revisions of "2024 AMC 10A Problems/Problem 15"
(→Solution 2) |
|||
Line 27: | Line 27: | ||
<cmath>y+x=1280</cmath> gives the solutions <math>y=641,x=639</math>. | <cmath>y+x=1280</cmath> gives the solutions <math>y=641,x=639</math>. | ||
− | Since we're only focusing on the units digit of <math>M=x^2-1213=y^2-3773</math>, then we only have to focus on the units digits of <math>x^2</math> and <math>1213</math> or <math>y^2</math> and <math>3773</math>. <math>1^2</math> and <math>9^2</math> give the same units digit, and <math>1213</math> and <math>3773</math> also have the same units digit. So the units digit of <math>M</math> is <math>10-|1-3|=\boxed{\text{(E) }8}</math> | + | Since we're only focusing on the units digit of <math>M=x^2-1213=y^2-3773</math>, then we only have to focus on the units digits of <math>x^2</math> and <math>1213</math> or <math>y^2</math> and <math>3773</math>. <math>1^2</math> and <math>9^2</math> give the same units digit, and <math>1213</math> and <math>3773</math> also have the same units digit. So the units digit of <math>M</math> is <math>10-|1-3|=\boxed{\text{(E) }8}</math> ~Tacos_are_yummy_1 |
+ | |||
+ | Note: I started creating this solution as the above solution already was. Since our solutions are awfully similar, I have no idea what to do. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:11, 8 November 2024
Contents
[hide]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
Finally, we get so the units digit of is
~MRENTHUSIASM
Solution 2
Let and . Subtracting the two equations will give Factoring the left hand side as a difference of squares gives The goal in the problem was to maximize , which therefore means maximizing both and and maximizing .
and are two factors of , therefore maximizing means making and as far as possible from each other. We start with , the most obvious choice that makes as large as possible, but solving and will give non-integer solutions, therefore this case does not work.
The next most obvious choice would be , which does give integer solutions to and . gives the solutions .
Since we're only focusing on the units digit of , then we only have to focus on the units digits of and or and . and give the same units digit, and and also have the same units digit. So the units digit of is ~Tacos_are_yummy_1
Note: I started creating this solution as the above solution already was. Since our solutions are awfully similar, I have no idea what to do.
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.