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

Latest revision as of 20:18, 9 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

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 a positive number, so that would be invalid! We see that - $30, -20, -15$ , and $-10$ are too far negative 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 less than $0$ . 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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 15: / 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>
-~ ~[[User:pog|pog]],~[[User:Mathkiddus|mathkiddus]]
+~[[User:pog|pog]],~[[User:Mathkiddus|mathkiddus]]
 == Solution 3 ==

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