Difference between revisions of "2010 AMC 12B Problems/Problem 14"
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Therefore, the answer is <math>\boxed{B}.</math> | Therefore, the answer is <math>\boxed{B}.</math> | ||
− | + | == Solution 3 == | |
Since <math>a + b \le M</math>, <math>d + e \le M</math>, and <math>c < b + c \le M</math>, we have that <math>2010 = a + b + c + d + e < 3M</math>. Hence, <math>M > 670</math>, or <math>M \ge 671</math>. | Since <math>a + b \le M</math>, <math>d + e \le M</math>, and <math>c < b + c \le M</math>, we have that <math>2010 = a + b + c + d + e < 3M</math>. Hence, <math>M > 670</math>, or <math>M \ge 671</math>. | ||
For the values <math>(a,b,c,d,e) = (669,1,670,1,669)</math>, <math>M = 671</math>, so the smallest possible value of <math>M</math> is <math>\boxed{671}</math>. The answer is (B). | For the values <math>(a,b,c,d,e) = (669,1,670,1,669)</math>, <math>M = 671</math>, so the smallest possible value of <math>M</math> is <math>\boxed{671}</math>. The answer is (B). | ||
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+ | ~ math31415926535 | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}} | {{AMC12 box|year=2010|num-b=13|num-a=15|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:33, 23 November 2020
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 2
First, note that, simply by pigeonhole, at least one of a, b, c, d, e is greater than or equal to so none of C, D, or E can be the answer. Thus, the answer is A or B. We will show that A is unattainable, leaving us with B as the only possible answer.
Assume WLOG that is the largest sum. So meaning Because we let we must have and Adding these inequalities gives But we just showed that which means that a contradiction because we are told that all the variables are positive.
Therefore, 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 |
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