2010 AMC 12B Problems/Problem 14
Contents
Problem 14
Let , , , , and be positive integers with and let be the largest of the sum , , and . What is the smallest possible value of ?
Solution 1
We want to try make , , , and as close as possible so that , the maximum of these, is smallest.
Notice that . In order to express as a sum of numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): or . We see that in both cases, the value of is , so the answer is .
Solution 3
Since , , and , we have that . Hence, , or .
For the values , , so the smallest possible value of is . The answer is (B).
~ math31415926535
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.