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

Revision as of 18:56, 8 November 2024

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

Video Solution 1 by Power Solve

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

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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 (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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. Error creating thumbnail: File missing

@@ Line 16: / Line 16: @@
 ~ pog, mathkiddus
+== Video Solution 1 by Power Solve ==
+https://youtu.be/j-37jvqzhrg?si=aggRgbnyZ3QjYwZF&t=806
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