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

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<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math>
 
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math>
−
 
−
==Solution 3==
 
−
let <cmath>m+1213=N^2</cmath>
 
−
<cmath>\Rightarrow m+3773=(N+a)^2</cmath>
 
−
It is obvious that <math>a\neq1</math> by parity
 
−
Thus, the minimum value of a 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>
 
−
=\fbox{(E) 8}
 
  
 
==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, \\
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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.  
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~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
 +
 +
==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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