2010 UNCO Math Contest II Problems/Problem 3

Problem

Suppose $r, s$ , and $t$ are three different positive integers and that their product is $48$ , i.e., $rst=48.$ What is the smallest possible value of the sum $r+s+t$ ?

Solution

$r+s+t$ is minimized when $r,s,t$ are farthest apart from each other. Therefore, $r=48,s=1,t=1$ solves the problem to be $48+1+1 = \boxed{\textbf{50}}$ .

2010 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All UNCO Math Contest Problems and Solutions

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2010 UNCO Math Contest II Problems/Problem 3

Problem

Solution

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