Difference between revisions of "2010 AMC 12B Problems/Problem 14"
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== Solution == | == Solution == | ||
| + | We want to try make <math>a+b</math>, <math>b+c</math>, <math>c+d</math>, and <math>d+e</math> as close as possible so that <math>M</math>, the maximum of these, if smallest. | ||
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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</math>. | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} | {{AMC12 box|year=2010|num-b=12|num-a=14|ab=B}} | ||
Revision as of 01:43, 18 January 2011
Problem 14
Let 
, 
, 
, 
, and 
be postive integers with 
and let 
be the largest of the sum 
, 
, 
and 
. What is the smallest possible value of ?
Solution
We want to try make , , , and as close as possible so that , the maximum of these, if 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 .
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 | |
