Difference between revisions of "2011 AMC 12B Problems/Problem 13"
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z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ | z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ | ||
4z + a + (a + b) + 9 &= 44\\ | 4z + a + (a + b) + 9 &= 44\\ | ||
| − | if \hspace{1cm} a &= 3 \\ | + | \text{if} \hspace{1cm} a &= 3 \\ |
a + b &= 4\\ | a + b &= 4\\ | ||
4z &= 44 - 9 - 3 - 4\\ | 4z &= 44 - 9 - 3 - 4\\ | ||
| Line 27: | Line 27: | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
| − | if \hspace{1cm} a &= 5\\ | + | \text{if} \hspace{1cm} a &= 5\\ |
a + b &= 6\\ | a + b &= 6\\ | ||
4z &= 44 - 9 - 5 - 6\\ | 4z &= 44 - 9 - 5 - 6\\ | ||
Revision as of 20:01, 5 October 2018
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 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
