Difference between revisions of "2011 AMC 12B Problems/Problem 13"
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<math>a+b</math>, and <math>b+c</math> then must be the remaining two numbers which are <math>4</math> and <math>6</math>. | <math>a+b</math>, and <math>b+c</math> then must be the remaining two numbers which are <math>4</math> and <math>6</math>. | ||
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>. | ||
+ | |||
Case 1 | Case 1 | ||
− | <math>(a,b,c)=(3,1,5) | + | <math> |
− | + | (a,b,c)=(3,1,5)\\ | |
− | + | x=w-5\\ | |
− | + | y=w-5-1\\ | |
− | + | x=w-5-1-3\\ | |
− | + | w+x+y+z=4w-20=44\\ | |
+ | w=16\\ </math> | ||
Case 2 | Case 2 | ||
− | <math>(a,b,c)=(5,1,3) | + | <math>(a,b,c)=(5,1,3)\\ |
− | + | x=w-3\\ | |
− | + | y=w-3-1\\ | |
− | + | x=w-3-1-5\\ | |
− | + | w+x+y+z=4w-16=44\\ | |
− | + | w=15</math> | |
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> |
Revision as of 21:09, 2 December 2017
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 .
Case 1
$(a,b,c)=(3,1,5)\\
x=w-5\\
y=w-5-1\\
x=w-5-1-3\\
w+x+y+z=4w-20=44\\
w=16\$ (Error compiling LaTeX. ! Missing $ inserted.)
Case 2
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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