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2024 AMC 10A Problems/Problem 7: Difference between revisions

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{{duplicate|[[2024 AMC 10A Problems/Problem 7|2024 AMC 10A #7]] and [[2024 AMC 12A Problems/Problem 6|2024 AMC 12A #6]]}}
==Problem==
==Problem==
The product of three integers is <math>60</math>. What is the least possible positive sum of the  
The product of three integers is <math>60</math>. What is the least possible positive sum of the  
Line 13: Line 15:
We have <math>abc = 60</math>. Let <math>a</math> be positive, and let <math>b</math> and <math>c</math> be negative. Then we need <math>a > |b + c|</math>. If <math>a = 6</math>, then <math>|b + c|</math> is at least <math>7</math>, so this doesn't work. If <math>a = 10</math>, then <math>(b,c) = (-6,-1)</math> works, giving <math>10 - 7 = \boxed{\textbf{(B) }3}</math>
We have <math>abc = 60</math>. Let <math>a</math> be positive, and let <math>b</math> and <math>c</math> be negative. Then we need <math>a > |b + c|</math>. If <math>a = 6</math>, then <math>|b + c|</math> is at least <math>7</math>, so this doesn't work. If <math>a = 10</math>, then <math>(b,c) = (-6,-1)</math> works, giving <math>10 - 7 = \boxed{\textbf{(B) }3}</math>


~ pog, mathkiddus
~[[User:pog|pog]],~[[User:Mathkiddus|mathkiddus]]
 
== Solution 3 ==
We can see that the most optimal solution would be <math>1</math> positive integer and <math>2</math> negative ones (as seen in solution 1). Let the three integers be <math>x</math>, <math>y</math>, and <math>z</math>, and let <math>x</math> be positive and <math>y</math> and <math>z</math> be negative. If we want the optimal solution, we want the negative numbers to be as large as possible. so the answer should be <math>-60 \cdot -1 \cdot 1</math>, where <math>-60 - 1 + 1 = -60</math>... right?
 
No! Our sum must be positive, so that would be invalid! We see that -<math>30, -20, -15</math>, and <math>-10</math> are too less to allow the sum to be positive. For example, <math>-15 = z. -15xy=60</math>, so <math>xy = -4</math>. For <math>xy</math> to be the most positive, we will have <math>4</math> and <math>-1</math>. Yet, <math>-15+4-1</math> is still negative. After <math>-10</math>, the next factor of <math>60</math> would be <math>-6</math>. if <math>z</math> = <math>-6</math>, <math>xy = -10</math>. This might be positive! Now, if we have <math>z = -6, y = -1</math>, and <math>x = 10, x + y + z = 3</math>. It cannot be smaller because <math>x = 5</math> and <math>y = -2</math> would result in <math>x + y + z</math> being negative. Therefore, our answer would be <math>\boxed{\textbf{(B) }3}</math>.
 
~Moonwatcher22
 
Tiny minor edits made by SuperVince1
 
== Video Solution by Apex Academy ==
 
https://youtu.be/bXyUErRWq5Q
 
==Video Solution(Don't be tricked!)==
https://youtu.be/l3VrUsZkv8I
 
== Video Solution by Math from my desk ==
 
https://www.youtube.com/watch?v=ptw2hYKoAIs&t=2s
 
== Video Solution (🚀 2 min solve 🚀) ==
 
https://youtu.be/u42_QzyO9zA
 
<i>~Education, the Study of Everything</i>
 
== Video Solution by Pi Academy ==
 
https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
 
== Video Solution 1 by Power Solve ==
https://youtu.be/j-37jvqzhrg?si=aggRgbnyZ3QjYwZF&t=806
 
== Video Solution by Daily Dose of Math ==
 
https://youtu.be/e8eL1l5os30
 
~Thesmartgreekmathdude
 
==Video Solution by SpreadTheMathLove==
https://www.youtube.com/watch?v=_o5zagJVe1U
 
==Video Solution by Dr. David==
https://youtu.be/5nRW7pb3ZXs
 
==Video solution by TheNeuralMathAcademy==
https://www.youtube.com/watch?v=4b_YLnyegtw&t=1033s


==See also==
==See Also==
{{AMC10 box|year=2024|ab=A|num-b=6|num-a=8}}
{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
{{AMC12 box|year=2024|ab=A|num-b=6|num-a=8}}
* [[AMC 10]]
* [[AMC 10 Problems and Solutions]]
* [[Mathematics competitions]]
* [[Mathematics competition resources]]
{{MAA Notice}}
{{MAA Notice}}

Latest revision as of 20:22, 2 November 2025

The following problem is from both the 2024 AMC 10A #7 and 2024 AMC 12A #6, so both problems redirect to this page.

Problem

The product of three integers is $60$. What is the least possible positive sum of the three integers?

$\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }13$

Solution 1

We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split $60$ into three factors and choose negativity. We notice that $10\cdot6\cdot1=10\cdot(-6)\cdot(-1)=60$, and trying other combinations does not yield lesser results so the answer is $10-6-1=\boxed{\textbf{(B) }3}$.

~eevee9406

Solution 2

We have $abc = 60$. Let $a$ be positive, and let $b$ and $c$ be negative. Then we need $a > |b + c|$. If $a = 6$, then $|b + c|$ is at least $7$, so this doesn't work. If $a = 10$, then $(b,c) = (-6,-1)$ works, giving $10 - 7 = \boxed{\textbf{(B) }3}$

~pog,~mathkiddus

Solution 3

We can see that the most optimal solution would be $1$ positive integer and $2$ negative ones (as seen in solution 1). Let the three integers be $x$, $y$, and $z$, and let $x$ be positive and $y$ and $z$ be negative. If we want the optimal solution, we want the negative numbers to be as large as possible. so the answer should be $-60 \cdot -1 \cdot 1$, where $-60 - 1 + 1 = -60$... right?

No! Our sum must be positive, so that would be invalid! We see that -$30, -20, -15$, and $-10$ are too less to allow the sum to be positive. For example, $-15 = z. -15xy=60$, so $xy = -4$. For $xy$ to be the most positive, we will have $4$ and $-1$. Yet, $-15+4-1$ is still negative. After $-10$, the next factor of $60$ would be $-6$. if $z$ = $-6$, $xy = -10$. This might be positive! Now, if we have $z = -6, y = -1$, and $x = 10, x + y + z = 3$. It cannot be smaller because $x = 5$ and $y = -2$ would result in $x + y + z$ being negative. Therefore, our answer would be $\boxed{\textbf{(B) }3}$.

~Moonwatcher22

Tiny minor edits made by SuperVince1

Video Solution by Apex Academy

https://youtu.be/bXyUErRWq5Q

Video Solution(Don't be tricked!)

https://youtu.be/l3VrUsZkv8I

Video Solution by Math from my desk

https://www.youtube.com/watch?v=ptw2hYKoAIs&t=2s

Video Solution (🚀 2 min solve 🚀)

https://youtu.be/u42_QzyO9zA

~Education, the Study of Everything

Video Solution by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 1 by Power Solve

https://youtu.be/j-37jvqzhrg?si=aggRgbnyZ3QjYwZF&t=806

Video Solution by Daily Dose of Math

https://youtu.be/e8eL1l5os30

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

Video Solution by Dr. David

https://youtu.be/5nRW7pb3ZXs

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=1033s

See Also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
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All AMC 10 Problems and Solutions

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