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

Revision as of 18:03, 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 18: / Line 18: @@
 == Solution 3 ==
-We can see that the most optimal solution would be <math>1</math> positive integer and <math>2</math> negetive 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?
+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 a positive number, so that would be invalid! We see that -<math>30, -20, -15</math>, and <math>-10</math> are too far negative 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 less than <math>0</math>. 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>.

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