2010 AMC 12B Problems/Problem 14
Problem
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 2
Since , , and , we have that . Hence, , or .
For the values , , so the smallest possible value of is . The answer is (B).
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Solution 3
Notice that only the sums of adjacent numbers matter. (For example, a & c could be extremely high, as long as b is relatively low.) Therefore creating "mountains" and "valleys" is the best way to lower the sum of adjacent numbers. We can do
1. (high, low, high, low, high)
or
2. (low, high, low, high, low)
In the extreme case that each "low" = , will be divided into either or numbers for cases 1 and 2, respectively. Obviously dividing by will yield a lower number, so we consider case 1.
Dividing by yields , or
()
However, all five numbers must be positive, and the closest we can get to this is
()
The lowest possible sum of two adjacent numbers then becomes , or .
See also
2010 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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