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

(Undo revision 245612 by Maa is stupid (talk))
(Tag: Undo)
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{{duplicate|[[2024 AMC 10A Problems/Problem 15|2024 AMC 10A #15]] and [[2024 AMC 12A Problems/Problem 9|2024 AMC 12A #9]]}}
 
==Problem==
 
==Problem==
Let <math>S</math> be a subset of <math>\{1, 2, 3, \cdots, 2024\}</math> such that the following two conditions hold:
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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>?
  
* If <math>x</math> and <math>y</math> are distinct elements of <math>S</math>, then <math>|x - y| > 2</math>.
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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>
* If <math>x</math> and <math>y</math> are distinct odd elements of <math>S</math>, then <math>|x - y| > 6</math>.
 
  
What is the maximum possible number of elements in <math>S</math>?
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==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>
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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
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<cmath>\begin{align*}
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Q+P&=1280, \\
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Q-P&=2,
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\end{align*}</cmath>
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from which <math>(P,Q)=(639,641).</math>
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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>
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~MRENTHUSIASM ~Tacos_are_yummy_1
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 +
==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>.
  
<math>\textbf{(A)}~436\qquad\textbf{(B)}~506\qquad\textbf{(C)}~608\qquad\textbf{(D)}~654\qquad\textbf{(E)}~675</math>
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~i_am_suk_at_math_2 (parity argument editing by Technodoggo)
  
==Video Solution by Scholars Foundation==
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==Solution 3==
https://www.youtube.com/watch?v=FKOqZau--5w&t=1s
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let <math>m+1213=N^2</math> <math>\Rightarrow m+3773=(N+a)^2</math>
  
==Solution 1==
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It is obvious that <math>a\neq1</math> by parity
All lists are organized in ascending order:
 
  
By listing out the smallest possible elements of subset <math>S,</math> we can find that subset <math>S</math> starts with <math>\{1, 4, 8, 11, 14, 18, 21, 24, 28, 31, \dots\}.</math> It is easily noticed that the elements of the subset "loop around" every 3 elements, specifically adding 10 each time. This means that there will be <math>2024/10</math> or <math>202R4</math> whole loops in the subset <math>S,</math> implying that there will be <math>202*3 = 606</math> elements in S. However, we have undercounted, as we did not count the remainder that resulted from <math>2024/10</math><math>.</math> With a remainder of <math>4,</math> we can fit <math>2</math> more elements into the subset <math>S,</math> namely <math>2021</math> and <math>2024,</math> resulting in a total of <math>606+2</math> or <math>\boxed{\textbf{(C) }608}</math> elements in subset <math>S</math>.
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Thus, the minimum value of <math>a</math> is 2
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Which gives us,
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<cmath>(N+a)^2-N^2=m+3773-(m+1213)</cmath>
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<cmath>4N+4=2560</cmath>
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<cmath>N=639</cmath>
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Plugging this back in,
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<cmath>m=N^2-1213 \space \mod \space 10</cmath>
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<cmath>m=8 \space \mod \space 10</cmath>
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Hence the answer <math>\boxed{\textbf{(E) }8}</math>.
  
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~lptoggled
  
NOTE:
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- trevian1(minor edit)
  
To prove that this is the best we can do, consider adding each element one by one, for the first element, say n. If n is greater than 2, we can choose n - 2 which is always better. Therefore, n = 1 or n = 2.
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==Solution 4==
  
If n = 2 was optimal, then choose it, then the set of usable numbers in <math>S</math> becomes 5 through 2024. We can transform the usable set of <math>S</math> to <math>Q</math> where <math>Q</math> contains the numbers 1 through 2020. Because we assumed n = 2 was optimal, we can choose n = 2 for the set <math>Q</math> too. Because every element in <math>Q</math> is 4 below the elements of <math>S</math>, choosing 2 in <math>Q</math> would mean choosing 6 in set <math>S</math>. By induction we see that our list would be {2,6,10,14,18,.....2022} which only gives 506 elements which is sub-optimal. Therefore, we can conclude that n = 1 is optimal, and we proceed as the solution above.
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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>.  
  
-weihou0
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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>
  
==Solution 2==
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~xHypotenuse
Notice that we can first place odd numbers, then place even numbers between each pair. We can start at <math>1</math> and continue from there. Realize that the smallest number <math>k</math> such that <math>kx+1</math> reproduces odd number is <math>8</math>. The next ones are <math>10, 12, 14</math>. We can proceed to find the number of numbers in this particular sequence. From the equation <math>8x+1=2023</math>, we get that <math>x \leq 252.875</math> works, so this means there is 253 solutions. Looking at <math>1,2,3,4,5,6,7,9</math> we can see that there could only be 1 possible number between each pair, yielding <math>252+253=505</math>. Then see that we can fit two more into the number count since the set <math>2017</math> to <math>2024</math> can fit two evens. Now this means <math>A</math> and <math>B</math> don’t work. Now test out <math>10x+1</math>. Using the same method, we get that <math>608</math> is the maximum number in the set. Everything above <math>x=10</math> doesn’t work, as we can split it down into smaller subgroups, so the answer is <math>\boxed{\textbf{(C) }608}</math>.
 
  
~EaZ_Shadow
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== Video Solution (⚡️ 4 min solve ⚡️)==
  
 +
https://youtu.be/YgJ23mepN0Q
  
== Video Solution by Power Solve ==
+
<i>~Education, the Study of Everything</i>
https://www.youtube.com/watch?v=NZ0SBMqeAfg
 
  
 
== Video Solution by Pi Academy ==
 
== Video Solution by Pi Academy ==
  
https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
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https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
 +
 
 +
== Video Solution 1 by Power Solve ==
 +
https://youtu.be/FvZVn0h3Yk4
  
 
==Video Solution by SpreadTheMathLove==
 
==Video Solution by SpreadTheMathLove==
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==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=14|num-a=16}}
 
{{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}}
 
{{MAA Notice}}

Revision as of 12:22, 21 March 2025

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

- trevian1(minor edit)

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 (⚡️ 4 min solve ⚡️)

https://youtu.be/YgJ23mepN0Q

~Education, the Study of Everything

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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png