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"

(→‎Solution 6)
(→‎Video Solution by Pi Academy)
 
(61 intermediate revisions by 24 users not shown)
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==
 
==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, 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
 +
<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==
 
==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.
 +
 
 +
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>.
  
−
oh and never gonna give you up
+
~i_am_suk_at_math_2 (parity argument editing by Technodoggo)
−
never gonna let you down
 
−
never gonna turn around and desert you
 
−
never gonna make you cry
 
−
never gonna say goodbye
 
−
never gonna tell a lie
 
−
and hurt you
 
  
 
==Solution 3==
 
==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>.
 +
 
 +
~lptoggled
  
 
==Solution 4==
 
==Solution 4==
−
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==
+
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>.
−
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>
+
 
 +
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
  
−
==Noitulos <math>\pi</math>==
+
==Video Solution by SpreadTheMathLove==
−
?????????? wtf bro <math>\textbf{(D)}</math>
+
https://www.youtube.com/watch?v=6SQ74nt3ynw
  
−
==Solution 6==
+
==See also==
−
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
+
{{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}}
−
<math>\boxed{\textbf{(D)}\frac{1}{0}}</math>
+
{{AMC12 box|year=2024|ab=A|num-b=8|num-a=10}}
 +
{{MAA Notice}}

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10A_Problems/Problem_15&oldid=237224"