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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
