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

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

@@ 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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