Difference between revisions of "2024 AMC 10A Problems/Problem 7"

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.

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Video Solution by Apex Academy
6 Video Solution(Don't be tricked!)
7 Video Solution by Math from my desk
8 Video Solution (🚀 2 min solve 🚀)
9 Video Solution by Pi Academy
10 Video Solution 1 by Power Solve
11 Video Solution by Daily Dose of Math
12 Video Solution by SpreadTheMathLove
13 Video Solution by Dr. David
14 Video solution by TheNeuralMathAcademy
15 See Also

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

@@ Line 25: / Line 25: @@
 Tiny minor edits made by SuperVince1
+== Video Solution by Apex Academy ==
+https://youtu.be/bXyUErRWq5Q
 ==Video Solution(Don't be tricked!)==

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