Difference between revisions of "2011 AMC 12B Problems/Problem 13"
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The ordering of <math>(a,b,c)</math> must be either <math>(3,1,5)</math> or <math>(5,1,3)</math>. | The ordering of <math>(a,b,c)</math> must be either <math>(3,1,5)</math> or <math>(5,1,3)</math>. | ||
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− | + | <cmath> | |
− | < | + | \begin{align*} |
− | + | z + (z + a) + (z + (a + b)) + (z + (a + b + c)) &= 4z + a + (a + b) + 9\\ | |
− | + | 4z + a + (a + b) + 9 &= 44\\ | |
− | < | + | \text{if} \hspace{1cm} a &= 3 \\ |
− | + | a + b &= 4\\ | |
− | + | 4z &= 44 - 9 - 3 - 4\\ | |
+ | z &= 7\\ | ||
+ | w &= 16\\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \text{if} \hspace{1cm} a &= 5\\ | ||
+ | a + b &= 6\\ | ||
+ | 4z &= 44 - 9 - 5 - 6\\ | ||
+ | z &= 6\\ | ||
+ | w &= 15\\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
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> | ||
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== 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 23:06, 25 December 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 |
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