Difference between revisions of "2024 AMC 10A Problems/Problem 7"
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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> | <math>\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }13</math> | ||
| − | ==Solution== | + | ==Solution 1== |
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>. | 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>. | ||
| + | == 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> | ||
| + | ~ pog, mathkiddus | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2024|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 15:50, 8 November 2024
Contents
Problem
The product of three integers is 60. What is the least possible positive sum of the three integers?
Solution 1
We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split 
into three factors and choose negativity. We notice that 
, and trying other combinations does not yield lesser results so the answer is 
.
Solution 2
We have 
. Let 
be positive, and let 
and 
be negative. Then we need 
. If 
, then 
is at least 
, so this doesn't work. If 
, then 
works, giving 
~ 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
