Difference between revisions of "2011 AMC 12B Problems/Problem 13"
(→Solution) |
|||
Line 7: | Line 7: | ||
Assume that <math>y-z=a, x-y=b, w-x=c.</math> | Assume that <math>y-z=a, x-y=b, w-x=c.</math> | ||
<math>w-z</math> results in the greatest pairwise difference, and thus it is <math>9</math>. | <math>w-z</math> results in the greatest pairwise difference, and thus it is <math>9</math>. | ||
− | + | This means <math>a+b+c=9</math>. <math>a,b,c</math> must be in the set <math>{1,3,4,5,6}</math>. | |
The only way for 3 numbers in the set to add up to 9 is if they are <math>1,3,5</math>. | The only way for 3 numbers in the set to add up to 9 is if they are <math>1,3,5</math>. | ||
<math>a+b</math>, and <math>b+c</math> then must be the remaining two numbers which are <math>4</math> and <math>6</math>. | <math>a+b</math>, and <math>b+c</math> then must be the remaining two numbers which are <math>4</math> and <math>6</math>. | ||
Line 29: | Line 29: | ||
The sum of the two w's is <math>15+16=31</math> <math>\boxed{B}</math> | The sum of the two w's is <math>15+16=31</math> <math>\boxed{B}</math> | ||
+ | |||
== See also == | == See also == | ||
{{AMC12 box|year=2011|ab=B|num-b=12|num-a=14}} | {{AMC12 box|year=2011|ab=B|num-b=12|num-a=14}} |
Revision as of 22:30, 26 June 2011
Problem
Brian writes down four integers whose sum is . The pairwise positive differences of these numbers are and . What is the sum of the possible values of ?
Solution
Assume that results in the greatest pairwise difference, and thus it is . This means . must be in the set . The only way for 3 numbers in the set to add up to 9 is if they are . , and then must be the remaining two numbers which are and . the ordering of must be either or .
Case 1
Case 2
The sum of the two w's is
See also
2011 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 |