Difference between revisions of "2024 AMC 10A Problems/Problem 7"
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| − | == Problem == | + | ==Problem== |
| − | The product of three integers is | + | The product of three integers is 60. What is the least possible positive sum of the |
| + | three integers? | ||
| − | <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== | ||
| + | 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>. | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=6|num-a=8}} | ||
| + | {{MAA Notice}} | ||
Revision as of 15:48, 8 November 2024
Problem
The product of three integers is 60. What is the least possible positive sum of the three integers?
Solution
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 
.
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. 
