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

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

@@ 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
+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==
-If you want to know where the king of ohio is you have to start mewing to skibidi and have a legendary mog battle with the all-knowing Ohio sigma known as master goonway who mews over the betas with l rizz and a level 0 on the rizz-o-meter. So stay skibidi and be the sigma over all betas so you can be the Ohio rizzler.
-also listen to Eminem clean
+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.
-oh and never gonna give you up
+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>.
-never gonna let you down
-never gonna turn around and desert you
+~i_am_suk_at_math_2 (parity argument editing by Technodoggo)
-never gonna make you cry
-never gonna say goodbye
-never gonna tell a lie
-and hurt you
 ==Solution 3==
-By the Ohio theorem, the answer is clearly not 6969420, or 42069. We can apply the Skibidi Slicers theorem to then, get the answer of <math>\boxed{\text{Baby Gronk - Livvy Dunne}}</math>
+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>.
-==Solution 4==
+~lptoggled
-Sigma ohio inequality states that <math>b\text{Sigma}^{a}\leq \sqrt{\text{Ohio}^{ab} \text{Mogging caseoh}} \leq +10000b \text{aura}</math>
-<math>\forall a,b \in \mathbb{SIGMA}</math>
-<math>\boxed{\textbf{(D)}+\infty \text{ aura}}</math>
 ==Solution 4==
-Using the brainrot theorem, we can see that the spheres are forming an exponential function, so we divide by the rizzler, and then multiply it by ohio. So the answer is <math>\boxed{D, 1434}</math>
-==Noitulos <math>\pi</math>==
+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>.
-?????????? wtf bro <math>\textbf{(D)}</math>
-==Solution 6==
+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>
-we do the thing (compose the gyatt theorem into the rizzler function) and it works, then apply fanum tax and tensor product <math>\otimes</math> with the mythical Ohio Grassman to yield
-<math>\boxed{\textbf{(D)}\frac{1}{0}}</math>
-==Solution 1434==
+~xHypotenuse
-May I have your attention, please?
-May I have your attention, please?
-Will the real Slim Shady please stand up?
-I repeat
-Will the real Slim Shady please stand up?
-We're gonna have a problem here
-Y'all act like you never seen a white person before
-Jaws all on the floor like Pam, like Tommy just burst in the door
-And started whoopin' her *ss worse than before
-They first were divorced, throwin' her over furniture (Agh)
-It's the return of the"Oh, wait, no way, you're kidding
-He didn't just say what I think he did, did he?"
-And Dr. Dre said
-Nothing, you idiots, Dr. Dre's dead, he's locked in my basement (Ha-ha)
-Feminist women love Eminem
-"Chicka-chicka-chicka, Slim Shady,I'm sick of him
-Look at him, walkin' around, grabbin' his you-know-what
-Flippin' the you-know-who", "Yeah, but he's so cute though"
-Yeah, I probably got a couple of screws up in my head loose
-But no worse than what's goin' on in your parents' bedrooms
-Sometimes I wanna get on TV and just let loose
-But can't, but it's cool for Tom Green to hump a dead moose
-"My bum is on your lips, my bum is on your lips"
-And if I'm lucky, you might just give it a little kiss
-And that's the message that we deliver to little kids
-And expect them not to know what a woman's clitoris is
-Of course, they're gonna know what intercourse is
-By the time they hit fourth gradethey've got the Discovery Channel, don't they?
-We ain't nothin' but mammals
-Well, some of us cannibals who cut other people open like cantaloupes
-But if we can hump dead animals and antelopes
-Then there's no reason that a man and another man can't elope
-But if you feel like I feel, I got the antidote
-Women, wave your pantyhose, sing the chorus, and it goes
-See Eminem Live
-Get tickets as low as $99
-You might also like
-Without Me
-Eminem
-Habits
-Eminem & White Gold
-But Daddy I Love Him
-Taylor Swift
-I'm Slim Shady, yes, I'm the real Shady
+== Video Solution by Pi Academy ==
-All you other Slim Shadys are just imitating
-So won't the real Slim Shady please stand up
-Please stand up, please stand up?
-'Cause I'm Slim Shady, yes, I'm the real Shady
-All you other Slim Shadys are just imitating
-So won't the real Slim Shady please stand up
-Please stand up, please stand up?
-Will Smith don't gotta cuss in his raps to sell records (Nope)
+https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
-Well, I do, so f**k him, and f**k you too
-You think I give a damn about a Grammy?
-Half of you critics can't even stomach me, let alone stand me
-"But Slim, what if you win? Wouldn't it be weird?"
-Why? So you guys could just lie to get me here?
-So you can sit me here next to Britney Spears?
-Yo, shit, Christina Aguilera better switch me chairs
-So I can sit next to Carson Daly and Fred Durst
-And hear 'em argue over who she gave head to first
-Little b**ch put me on blast on MTV
-"Yeah, he's cute, but I think he's married to Kim, hee-hee"
-I should download her audio on MP3
-And show the whole world how you gave Eminem VD (Agh)
-I'm sick of you little girl and boy groups, all you do is annoy me
-So I have been sent here to destroy you
-And there's a million of us just like me
-Who cuss like me, who just don't give a f**k like me
-Who dress like me, walk, talk and act like me
-And just might be the next best thing, but not quite me
-'Cause I'm Slim Shady, yes, I'm the real Shady
-All you other Slim Shadys are just imitating
-So won't the real Slim Shady please stand up
-Please stand up, please stand up?
-'Cause I'm Slim Shady, yes, I'm the real Shady
-All you other Slim Shadys are just imitating
-So won't the real Slim Shady please stand up
-Please stand up, please stand up?
-I'm like a head trip to listen to
+== Video Solution 1 by Power Solve ==
-'Cause I'm only givin' you things you joke about with your friends inside your livin' room
+https://youtu.be/FvZVn0h3Yk4
-The only difference is I got the balls to say it in front of y'all
-And I don't gotta be false or sugarcoat it at all
-I just get on the mic and spit it
-And whether you like to admit it (Err), I just s**t it
-Better than ninety percent of you rappers out can
-Then you wonder, "How can kids eat up these albums like Valiums?"
-It's funny, 'cause at the rate I'm goin', when I'm thirty
-I'll be the only person in the nursin' home flirting
-Pinchin' nurse's *sses when I'm jacking off with Jergens
-And I'm jerking, but this whole bag of Viagra isn't working
-And every single person is a Slim Shady lurkin'
-He could be working at Burger King, spittin' on your onion rings (Ch, puh)
-Or in the parkin' lot, circling, screaming, "I don't give a f**k!"
-With his windows down and his system up
-So will the real Shady please stand up
-And put one of those fingers on each hand up?
-And be proud to be out of your mind and out of control
-And one more time, loud as you can, how does it go?
-I'm Slim Shady, yes, I'm the real Shady
+==Video Solution by SpreadTheMathLove==
-All you other Slim Shadys are just imitating
+https://www.youtube.com/watch?v=6SQ74nt3ynw
-So won't the real Slim Shady please stand up
-Please stand up, please stand up?
-'Cause I'm Slim Shady, yes, I'm the real Shady
-All you other Slim Shadys are just imitating
-So won't the real Slim Shady please stand up
-Please stand up, please stand up?
-'Cause I'm Slim Shady, yes, I'm the real Shady
-All you other Slim Shadys are just imitating
-So won't the real Slim Shady please stand up
-Please stand up, please stand up?
-'Cause I'm Slim Shady, yes, I'm the real Shady
-All you other Slim Shadys are just imitating
-So won't the real Slim Shady please stand up
-Please stand up, please stand up?
-Ha-ha
+==See also==
-I guess there's a Slim Shady in all of us
+{{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}}
-F**k it, let's all stand up
+{{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
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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