Difference between revisions of "2010 UNCO Math Contest II Problems/Problem 3"

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== Solution ==
 
== Solution ==
<math>r+s+t</math> is minimized when <math>r,s,t</math> are farthest apart from each other. Therefore, <math>r=48,s=1,t=1</math> solves the problem to be <math>48+1+1 = \boxed{\textbf{50}}</math>.
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The solution rst must factor <math>48=2^{4}\cdot{3}</math> and the numbers must be DISTINCT. <math>3+4+4=11</math> fails the distinct test so <math>2+4+6=12</math> is the solution.
  
 
== See also ==
 
== See also ==

Latest revision as of 02:46, 13 January 2019

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

The solution rst must factor $48=2^{4}\cdot{3}$ and the numbers must be DISTINCT. $3+4+4=11$ fails the distinct test so $2+4+6=12$ is the solution.

See also

2010 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions