2011 AMC 12B Problems/Problem 13
Contents
[hide]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 
.
The sum of the two w's is 
Solution 2
Let the four numbers be 
, 
, 
, and 
. We know that 
must be because that's the greatest difference. So we have , , , and 
. The 6 possible differences are 
, 
, , 
, 
, and 
. We are given that the differences are 1, 3, 4, 5, 6, 9. and and and add to 9 which means they have to be 4, 5 and 3,6 or vice versa. Which leaves 
. That means has to equal . So an and b have to be 3,4, 4,5, or 5,6. For 3,4, we have , 
, 
, and . 
. 
. 
. 
. Now for 4,5, notice that it doesn't work. The differences are 4, 5, 9, 1, 4, 5. We are missing 6 and 3. For 5,6, it's , 
, 
, and . Check that the differences work; they do. We have 
. 
. 
. 
. Therefore our answer is 
.
~MC413551
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 | |
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