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>. | ||
== See also == | == See also == | ||
Revision as of 22:14, 9 February 2015
Problem
Suppose 
, and 
are three different positive integers and that their product is 
, i.e., 
What is the smallest possible value of the sum 
?
Solution
is minimized when 
are farthest apart from each other. Therefore, 
solves the problem to be 
.
See also
| 2010 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| 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 | ||
