Difference between revisions of "2010 AMC 12B Problems/Problem 14"
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Notice that <math>2010=670+670+670</math>. In order to express <math>2010</math> as a sum of <math>5</math> 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): <math>2010=670+1+670+1+668</math> or <math>2010=670+1+669+1+669</math>. We see that in both cases, the value of <math>M</math> is <math>671</math>, so the answer is <math>671 \Rightarrow \boxed{B}</math>. | Notice that <math>2010=670+670+670</math>. In order to express <math>2010</math> as a sum of <math>5</math> 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): <math>2010=670+1+670+1+668</math> or <math>2010=670+1+669+1+669</math>. We see that in both cases, the value of <math>M</math> is <math>671</math>, so the answer is <math>671 \Rightarrow \boxed{B}</math>. | ||
− | == Solution | + | == Solution 2 == |
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>. | ||
Revision as of 00:30, 15 January 2021
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 2
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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