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

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==Solution 1==
 
==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{(B)3}</math>.
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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
 
== Solution 2==
 
== Solution 2==
 
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>
 
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>

Revision as of 15:53, 8 November 2024

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

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

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