Difference between revisions of "2013 UNCO Math Contest II Problems/Problem 5"

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== Solution ==
 
== Solution ==
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Intuitively, we want all the numbers to be as close as possible to <math>\tfrac{17}4</math>, or <math>4.5</math>. Thus, we get the numbers <math>2</math>, <math>4</math>, <math>5</math>, and <math>6</math>. These multiply up to <math>2\cdot4\cdot5\cdot6 = \boxed{240}</math>.
  
 
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~pineconee
 
== See Also ==
 
== See Also ==
 
{{UNCO Math Contest box|n=II|year=2013|num-b=4|num-a=6}}
 
{{UNCO Math Contest box|n=II|year=2013|num-b=4|num-a=6}}
  
 
[[Category:Intermediate Algebra Problems]]
 
[[Category:Intermediate Algebra Problems]]

Latest revision as of 15:04, 30 April 2022

Problem

If the sum of distinct positive integers is $17$, find the largest possible value of their product. Give both a set of positive integers and their product. Remember to consider only sums of distinct numbers, and not $3+7+7$ or $2+3+4+4+4$, etc., which have repeated terms. You need not justify your answer on this question.

$\begin{array}{|c|c|c|c|} \hline \text{EXAMPLE: }& \text{Distinct Integers: }{2, 3, 4, 8} & \text{Their Sum: }2+3+4+8=17 & \text{Their Product: }2 \times 3\times 4\times 8=192 \\ \hline \end{array}$

Solution

Intuitively, we want all the numbers to be as close as possible to $\tfrac{17}4$, or $4.5$. Thus, we get the numbers $2$, $4$, $5$, and $6$. These multiply up to $2\cdot4\cdot5\cdot6 = \boxed{240}$.

~pineconee

See Also

2013 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions