Difference between revisions of "2011 AMC 10B Problems/Problem 21"
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Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6,</math> and <math>9</math>. What is the sum of the possible values for <math>w</math>? | Brian writes down four integers <math>w > x > y > z</math> whose sum is <math>44</math>. The pairwise positive differences of these numbers are <math>1, 3, 4, 5, 6,</math> and <math>9</math>. What is the sum of the possible values for <math>w</math>? | ||
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w&=15 \end{align*}</cmath> | w&=15 \end{align*}</cmath> | ||
The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math> | The sum of the possible values of <math>w</math> is <math>16+15 = \boxed{\textbf{(B) }31}</math> | ||
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+ | == See Also== | ||
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+ | {{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}} |
Revision as of 19:51, 4 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 for ?
Solution
The largest difference, must be between and
The smallest difference, must be directly between two integers. This also means the differences directly between the other two should add up to The only remaining differences that would make this possible are and However, those two differences can't be right next to each other because they would make a difference of This means must be the difference between and We can express the possible configurations as the lines.
If we look at the first number line, you can express as as and as Since the sum of all these integers equal , You can do something similar to this with the second number line to find the other possible value of The sum of the possible values of is
See Also
2011 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 10 Problems and Solutions |