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

(Created page with "The official test date is in November 2024. A hacker has sent this question (2024 AMC 10A Problem 7) to a random pixel on a 1440x800 pixel computer screen for exactly 7.6230 s...")
 
 
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The official test date is in November 2024. A hacker has sent this question (2024 AMC 10A Problem 7) to a random pixel on a 1440x800 pixel computer screen for exactly 7.6230 seconds. What is the probability that, by randomly clicking your screen once at a random time of the day, that you would click the pixel with the question?
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{{duplicate|[[2024 AMC 10A Problems/Problem 7|2024 AMC 10A #7]] and [[2024 AMC 12A Problems/Problem 6|2024 AMC 12A #6]]}}
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==Problem==
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The product of three integers is <math>60</math>. What is the least possible positive sum of the
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three integers?
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<math>\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }13</math>
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==Solution 1==
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We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split <math>60</math> into three factors and choose negativity. We notice that <math>10\cdot6\cdot1=10\cdot(-6)\cdot(-1)=60</math>, and trying other combinations does not yield lesser results so the answer is <math>10-6-1=\boxed{\textbf{(B) }3}</math>.
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~eevee9406
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== Solution 2==
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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>
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~[[User:pog|pog]],~[[User:Mathkiddus|mathkiddus]]
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== Solution 3 ==
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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?
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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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~Moonwatcher22
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== Video Solution by Pi Academy ==
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https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
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== Video Solution 1 by Power Solve ==
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https://youtu.be/j-37jvqzhrg?si=aggRgbnyZ3QjYwZF&t=806
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/e8eL1l5os30
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~Thesmartgreekmathdude
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=_o5zagJVe1U
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==See also==
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{{AMC10 box|year=2024|ab=A|num-b=6|num-a=8}}
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{{AMC12 box|year=2024|ab=A|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 21:44, 13 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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

See also

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

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