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

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==Solution 1==
 
==Solution 1==
  
−
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>  
+
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> so we maximize <math>Q+P</math> and minimize <math>Q-P.</math> It follows that
+
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*}
 
<cmath>\begin{align*}
 
Q+P&=1280, \\
 
Q+P&=1280, \\
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
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It is obvious that <math>a\neq1</math> by parity
 
It is obvious that <math>a\neq1</math> by parity
  
−
Thus, the minimum value of a is 2
+
Thus, the minimum value of <math>a</math> is 2
 
Which gives us,
 
Which gives us,
 
<cmath>(N+a)^2-N^2=m+3773-m+1213</cmath>
 
<cmath>(N+a)^2-N^2=m+3773-m+1213</cmath>
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<cmath>m=N^2-1213 \space \mod \space 10</cmath>
 
<cmath>m=N^2-1213 \space \mod \space 10</cmath>
 
<cmath>m=8 \space \mod \space 10</cmath>
 
<cmath>m=8 \space \mod \space 10</cmath>
−
Hence the answer <math> \fbox{(E) 8}</math>
+
Hence the answer <math>\boxed{\textbf{(E) }8}</math>.
  
 
~lptoggled
 
~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
 +
 +
 +
== 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://www.youtube.com/watch?v=6SQ74nt3ynw
  
 
==See also==
 
==See also==

Latest 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.

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 (or else $Q$ and $P$ would not be integers), and $Q+P>Q-P.$

We wish to maximize both $P$ and $Q$ (Because we want to maximize $M$), 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

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 by Pi Academy

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


Video Solution 1 by Power Solve

https://youtu.be/FvZVn0h3Yk4

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

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