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

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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==
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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>?
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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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<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8</math>
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* If <math>x</math> and <math>y</math> are distinct elements of <math>S</math>, then <math>|x - y| > 2</math>.
 +
* If <math>x</math> and <math>y</math> are distinct odd elements of <math>S</math>, then <math>|x - y| > 6</math>.
  
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==Solution 1==
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What is the maximum possible number of elements in <math>S</math>?
  
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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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<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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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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==Video Solution by Scholars Foundation==
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<cmath>\begin{align*}
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https://www.youtube.com/watch?v=FKOqZau--5w&t=1s
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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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==Solution 1==
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All lists are organized in ascending order:
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~MRENTHUSIASM ~Tacos_are_yummy_1
 
  
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==Solution 2==
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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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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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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>.
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NOTE:
  
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~i_am_suk_at_math_2 (parity argument editing by Technodoggo)
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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 3==
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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> <math>\Rightarrow m+3773=(N+a)^2</math>
 
  
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It is obvious that <math>a\neq1</math> by parity
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-weihou0
  
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Thus, the minimum value of <math>a</math> is 2
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==Solution 2==
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Which gives us,
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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>.
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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
 
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- trevian1(minor edit)
 
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==Solution 4==
 
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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>.  
 
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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>
 
  
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~xHypotenuse
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~EaZ_Shadow
  
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== Video Solution (⚡️ 4 min solve ⚡️)==
 
  
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https://youtu.be/YgJ23mepN0Q
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== Video Solution by Power Solve ==
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https://www.youtube.com/watch?v=NZ0SBMqeAfg
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<i>~Education, the Study of Everything</i>
 
  
 
== Video Solution by Pi Academy ==
 
== Video Solution by Pi Academy ==
  
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https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM
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https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
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== Video Solution 1 by Power Solve ==
 
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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}}
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{{AMC12 box|year=2024|ab=A|num-b=8|num-a=10}}
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:28, 20 March 2025

Problem

Let $S$ be a subset of $\{1, 2, 3, \cdots, 2024\}$ such that the following two conditions hold:

  • If $x$ and $y$ are distinct elements of $S$, then $|x - y| > 2$.
  • If $x$ and $y$ are distinct odd elements of $S$, then $|x - y| > 6$.

What is the maximum possible number of elements in $S$?

$\textbf{(A)}~436\qquad\textbf{(B)}~506\qquad\textbf{(C)}~608\qquad\textbf{(D)}~654\qquad\textbf{(E)}~675$

Video Solution by Scholars Foundation

https://www.youtube.com/watch?v=FKOqZau--5w&t=1s

Solution 1

All lists are organized in ascending order:

By listing out the smallest possible elements of subset $S,$ we can find that subset $S$ starts with $\{1, 4, 8, 11, 14, 18, 21, 24, 28, 31, \dots\}.$ 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 $2024/10$ or $202R4$ whole loops in the subset $S,$ implying that there will be $202*3 = 606$ elements in S. However, we have undercounted, as we did not count the remainder that resulted from $2024/10$$.$ With a remainder of $4,$ we can fit $2$ more elements into the subset $S,$ namely $2021$ and $2024,$ resulting in a total of $606+2$ or $\boxed{\textbf{(C) }608}$ elements in subset $S$.


NOTE:

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.

If n = 2 was optimal, then choose it, then the set of usable numbers in $S$ becomes 5 through 2024. We can transform the usable set of $S$ to $Q$ where $Q$ contains the numbers 1 through 2020. Because we assumed n = 2 was optimal, we can choose n = 2 for the set $Q$ too. Because every element in $Q$ is 4 below the elements of $S$, choosing 2 in $Q$ would mean choosing 6 in set $S$. 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.

-weihou0

Solution 2

Notice that we can first place odd numbers, then place even numbers between each pair. We can start at $1$ and continue from there. Realize that the smallest number $k$ such that $kx+1$ reproduces odd number is $8$. The next ones are $10, 12, 14$. We can proceed to find the number of numbers in this particular sequence. From the equation $8x+1=2023$, we get that $x \leq 252.875$ works, so this means there is 253 solutions. Looking at $1,2,3,4,5,6,7,9$ we can see that there could only be 1 possible number between each pair, yielding $252+253=505$. Then see that we can fit two more into the number count since the set $2017$ to $2024$ can fit two evens. Now this means $A$ and $B$ don’t work. Now test out $10x+1$. Using the same method, we get that $608$ is the maximum number in the set. Everything above $x=10$ doesn’t work, as we can split it down into smaller subgroups, so the answer is $\boxed{\textbf{(C) }608}$.

~EaZ_Shadow


Video Solution by Power Solve

https://www.youtube.com/watch?v=NZ0SBMqeAfg

Video Solution by Pi Academy

https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE

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

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

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