Difference between revisions of "2011 AMC 12B Problems/Problem 13"
(→Solution) |
(→Solution) |
||
| Line 10: | Line 10: | ||
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>. | ||
| − | + | The ordering of <math>(a,b,c)</math> must be either <math>(3,1,5)</math> or <math>(5,1,3)</math>. | |
Case 1 | Case 1 | ||
Revision as of 21:30, 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 | |
