Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(sol 2's process is slightly better than 2a's, so combining the justification for 2a and the process for 2 makes sense =D also, I wonder who the reader of this message is! 1434 hehehehhehehhe he he he eh eh eh e)
 
(29 intermediate revisions by 17 users not shown)
Line 16: Line 16:
 
from which <math>(P,Q)=(639,641).</math>
 
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>
+
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
 
~MRENTHUSIASM ~Tacos_are_yummy_1
  
−
==Solution 2 (not rigorously proven)==
+
==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.  
 
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.  
Line 27: Line 27:
  
 
~i_am_suk_at_math_2 (parity argument editing by Technodoggo)
 
~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
 +
 +
- trevian1(minor edit)
 +
 +
==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
 +
 +
==Video Solution(Fast! About ⚡️ 3 min solve! ⚡️)==
 +
https://youtu.be/l3VrUsZkv8I
 +
 +
~MC
 +
== Video Solution (⚡️ 4 min solve ⚡️)==
 +
 +
https://youtu.be/YgJ23mepN0Q
 +
 +
<i>~Education, the Study of Everything</i>
 +
 +
== Video Solution by Pi Academy ==
 +
 +
https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
 +
 +
== Video Solution 1 by Power Solve ==
 +
https://youtu.be/FvZVn0h3Yk4
 +
 +
==Video Solution by SpreadTheMathLove==
 +
https://youtu.be/CmIPAvwtWLA?si=ZCv3ypdDmCaV-aX3
 +
 +
==Video Solution by Dr. David==
 +
https://youtu.be/XLoetj5obYE
  
 
==See also==
 
==See also==

Latest revision as of 11:47, 11 June 2025

The following problem is from both the 2024 AMC 10A #15 and 2024 AMC 12A #9, so both problems redirect to this page.

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 1

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 $(P,Q)=(639,641).$

Finally, we get $M=P^2-1213=Q^2-3773\equiv1-3\equiv8\pmod{10},$ so the units digit of $M$ is $\boxed{\textbf{(E) }8}.$

~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 $M+1213$ and $M+3773$ (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 $M+1213$ and $M+3773$ have one square in between them.

Let the square between $M+1213$ and $M+3773$ be $x^2$. So, we have $M+1213 = (x-1)^2$ and $M+3773 = (x+1)^2$. Subtracting the two, we have $(M+3773)-(M+1213) = (x+1)^2 - (x-1)^2$, which yields $2560 = 4x$, which leads to $x = 640$. Therefore, the two squares are $639^2$ and $641^2$, which both have units digit $1$. Since both $1213$ and $3773$ have units digit $3$, $M$ will have units digit $\boxed{\textbf{(E) }8}$.

~i_am_suk_at_math_2 (parity argument editing by Technodoggo)

Solution 3

Let $m+1213=N^2$ $\Rightarrow m+3773=(N+a)^2$

It is obvious that $a\neq1$ by parity

Thus, the minimum value of $a$ is 2 Which gives us, \[(N+a)^2-N^2=m+3773-(m+1213)\] \[4N+4=2560\] \[N=639\] Plugging this back in, \[m=N^2-1213 \space \mod \space 10\] \[m=8 \space \mod \space 10\] Hence the answer $\boxed{\textbf{(E) }8}$.

~lptoggled

- trevian1(minor edit)

Solution 4

Let $M+1213=n^2$ and $M+3773=(n+1)^2$ for some positive integer $n$. We do this because, in order to maximize $M$, 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 $2n+1=2560$; impossible. Then we try $M+3773=(n+2)^2$. Now we would have $4n+4=2560$ which indeed works! $n=639$.

Finally, we get $M=n^2-1213$ so the units digit of $M$ is $11-3=\boxed{\textbf{(E) }8}.$

~xHypotenuse

Video Solution(Fast! About ⚡️ 3 min solve! ⚡️)

https://youtu.be/l3VrUsZkv8I

~MC

Video Solution (⚡️ 4 min solve ⚡️)

https://youtu.be/YgJ23mepN0Q

~Education, the Study of Everything

Video Solution by Pi Academy

https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM

Video Solution 1 by Power Solve

https://youtu.be/FvZVn0h3Yk4

Video Solution by SpreadTheMathLove

https://youtu.be/CmIPAvwtWLA?si=ZCv3ypdDmCaV-aX3

Video Solution by Dr. David

https://youtu.be/XLoetj5obYE

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_15&oldid=250998"