Difference between revisions of "2024 AMC 10A Problems/Problem 15"
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− | What is the | + | {{duplicate|[[2024 AMC 10A Problems/Problem 15|2024 AMC 10A #15]] and [[2024 AMC 12A Problems/Problem 9|2024 AMC 12A #9]]}} |
+ | ==Problem== | ||
+ | Let <math>M</math> be the greatest integer such that both <math>M+1213</math> and <math>M+3773</math> are perfect squares. What is the units digit of <math>M</math>? | ||
− | + | <math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math> | |
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==Solution 1== | ==Solution 1== | ||
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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 (or else <math>Q</math> and <math>P</math> would not be integers), and <math>Q+P>Q-P.</math> | |
− | + | We wish to maximize both <math>P</math> and <math>Q</math> (Because we want to maximize <math>M</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> | ||
− | + | Finally, we get <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 ~Tacos_are_yummy_1 | |
− | + | ==Solution 2== | |
− | + | Ideally, we would like for the two squares to be as close as possible. The best case is that they are consecutive squares (no square numbers in between them); however, since <math>M+1213</math> and <math>M+3773</math> (and thus their squares) have the same parity, they cannot be consecutive squares (which are always opposite parities). The best we can do is that <math>M+1213</math> and <math>M+3773</math> have one square in between them. | |
− | + | Let the square between <math>M+1213</math> and <math>M+3773</math> be <math>x^2</math>. So, we have <math>M+1213 = (x-1)^2</math> and <math>M+3773 = (x+1)^2</math>. Subtracting the two, we have <math>(M+3773)-(M+1213) = (x+1)^2 - (x-1)^2</math>, which yields <math>2560 = 4x</math>, which leads to <math>x = 640</math>. Therefore, the two squares are <math>639^2</math> and <math>641^2</math>, which both have units digit <math>1</math>. Since both <math>1213</math> and <math>3773</math> have units digit <math>3</math>, <math>M</math> will have units digit <math>\boxed{\textbf{(E) }8}</math>. | |
− | + | ~i_am_suk_at_math_2 (parity argument editing by Technodoggo) | |
− | + | ==Solution 3== | |
+ | let <math>m+1213=N^2</math> <math>\Rightarrow m+3773=(N+a)^2</math> | ||
− | + | It is obvious that <math>a\neq1</math> by parity | |
− | + | Thus, the minimum value of <math>a</math> is 2 | |
+ | Which gives us, | ||
+ | <cmath>(N+a)^2-N^2=m+3773-m+1213</cmath> | ||
+ | <cmath>4N+4=2560</cmath> | ||
+ | <cmath>N=639</cmath> | ||
+ | Plugging this back in, | ||
+ | <cmath>m=N^2-1213 \space \mod \space 10</cmath> | ||
+ | <cmath>m=8 \space \mod \space 10</cmath> | ||
+ | Hence the answer <math>\boxed{\textbf{(E) }8}</math>. | ||
− | + | ~lptoggled | |
− | + | ==Solution 4== | |
+ | Let <math>M+1213=n^2</math> and <math>M+3773=(n+1)^2</math> for some positive integer <math>n</math>. We do this because, in order to maximize <math>M</math>, the perfect squares need to be as close to each other as possible. We find that this configuration doesn't work, as when we subtract the equations, we have <math>2n+1=2560</math>; impossible. Then we try <math>M+3773=(n+2)^2</math>. Now we would have <math>4n+4=2560</math> which indeed works! <math>n=639</math>. | ||
− | + | Finally, we get <math>M=n^2-1213</math> so the units digit of <math>M</math> is <math>11-3=\boxed{\textbf{(E) }8}.</math> | |
− | + | ~xHypotenuse | |
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− | ==Solution | + | == Video Solution by Pi Academy == |
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− | + | https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM | |
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− | + | == Video Solution 1 by Power Solve == | |
− | + | https://youtu.be/FvZVn0h3Yk4 | |
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+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=6SQ74nt3ynw | ||
− | + | ==See also== | |
− | + | {{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}} | |
− | + | {{AMC12 box|year=2024|ab=A|num-b=8|num-a=10}} | |
− | + | {{MAA Notice}} |
Revision as of 21:48, 13 November 2024
- The following problem is from both the 2024 AMC 10A #15 and 2024 AMC 12A #9, so both problems redirect to this page.
Contents
[hide]Problem
Let be the greatest integer such that both and are perfect squares. What is the units digit of ?
Solution 1
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 (or else and would not be integers), and
We wish to maximize both and (Because we want to maximize ), so we maximize and minimize It follows that from which
Finally, we get so the units digit of is
~MRENTHUSIASM ~Tacos_are_yummy_1
Solution 2
Ideally, we would like for the two squares to be as close as possible. The best case is that they are consecutive squares (no square numbers in between them); however, since and (and thus their squares) have the same parity, they cannot be consecutive squares (which are always opposite parities). The best we can do is that and have one square in between them.
Let the square between and be . So, we have and . Subtracting the two, we have , which yields , which leads to . Therefore, the two squares are and , which both have units digit . Since both and have units digit , will have units digit .
~i_am_suk_at_math_2 (parity argument editing by Technodoggo)
Solution 3
let
It is obvious that by parity
Thus, the minimum value of is 2 Which gives us, Plugging this back in, Hence the answer .
~lptoggled
Solution 4
Let and for some positive integer . We do this because, in order to maximize , the perfect squares need to be as close to each other as possible. We find that this configuration doesn't work, as when we subtract the equations, we have ; impossible. Then we try . Now we would have which indeed works! .
Finally, we get so the units digit of is
~xHypotenuse
Video Solution by Pi Academy
https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
Video Solution 1 by Power Solve
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=6SQ74nt3ynw
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 |
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.