2011 AMC 12B Problems/Problem 13

Revision as of 22:29, 26 June 2011 by Justinxu (talk | contribs)

Problem

Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6$ and $9$. What is the sum of the possible values of $w$?

$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 31 \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 62 \qquad \textbf{(E)}\ 93$

Solution

Assume that $y-z=a, x-y=b, w-x=c.$ $w-z$ results in the greatest pairwise difference, and thus it is $9$. which means $a+b+c=9$. $a,b,c$ must be in the set ${1,3,4,5,6}$. The only way for 3 numbers in the set to add up to 9 is if they are $1,3,5$. $a+b$, and $b+c$ then must be the remaining two numbers which are $4$ and $6$. the ordering of $(a,b,c)$ must be either $(3,1,5)$ or $(5,1,3)$.

Case 1 $(a,b,c)=(3,1,5)$ $x=w-5$ $y=w-5-1$ $x=w-5-1-3$ $w+x+y+z=4w-20=44$ $w=16$

Case 2 $(a,b,c)=(5,1,3)$ $x=w-3$ $y=w-3-1$ $x=w-3-1-5$ $w+x+y+z=4w-16=44$ $w=15$

The sum of the two w's is $15+16=31$ $\boxed{B}$

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS