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

Latest revision as of 11:29, 18 December 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, 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

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

@@ Line 1: / Line 1: @@
-Bob, the skibidi rizzler and mewer from Ohio just realized that he loves the song: "Sigma sigma on the wall", so he buys 13 mewing griddies to launch a nuke at all the fanum taxers that keep screaming for more spheres. How many skibidi toilets will it take to stop these ermos?
+{{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>
 ==Solution 1==
-because skibidi toilets are skibidi, it I will take sigma amount of skibidi toilets to stop the ermos but the fanum taxers will keep screaming for more spheres so you need sigma skibidi ermos to stop the fanum taxers
-==Noitulos 99^8==
+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>
-?????????? <math>\textbf{(D)}</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
+<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
+== 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==
+{{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}}

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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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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