Difference between revisions of "2011 AMC 12B Problems/Problem 13"
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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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+ | == Solution 2 == | ||
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+ | Let the four numbers be <math>z</math>, <math>z+a</math>, <math>z+b</math>, and <math>z+c</math>. We know that <math>c</math> must be <math>9</math> because that's the greatest difference. So we have <math>z</math>, <math>z+a</math>, <math>z+b</math>, and <math>z+9</math>. The 6 possible differences are <math>a</math>, <math>b</math>, <math>9</math>, <math>b-a</math>, <math>9-a</math>, and <math>9-b</math>. We are given that the differences are 1, 3, 4, 5, 6, 9. <math>a</math> and <math>9-a</math> and <math>b</math> and <math>9-b</math> add to 9 which means they have to be 4, 5 and 3,6 or vice versa. Which leaves <math>1</math>. That means <math>b-a</math> has to equal <math>1</math>. So an and b have to be 3,4, 4,5, or 5,6. For 3,4, we have <math>z</math>, <math>z+3</math>, <math>z+4</math>, and <math>z+9</math>. <math>4z+16=44</math>. <math>4z=28</math>. <math>z=7</math>. <math>w=7+9=16</math>. 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 <math>z</math>, <math>z+5</math>, <math>z+6</math>, and <math>z+9</math>. Check that the differences work; they do. We have <math>4z+20=44</math>. <math>4z=24</math>. <math>z=6</math>. <math>w=6+9=15</math>. Therefore our answer is <math>15+16=\boxed{31}</math>. | ||
+ | ~MC413551 | ||
== 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}} | {{MAA Notice}} |
Latest revision as of 08:55, 8 August 2023
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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