Difference between revisions of "2011 AMC 10B Problems/Problem 21"
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== Solution 3 == | == Solution 3 == | ||
| − | Let w - x = a, w - y = b, w - z = c. As above, we know that c = 9. Thus, a < b < c. | + | Let <math>w - x = a</math>, <math>w - y = b</math>, <math>w - z = c</math>. As above, we know that <math>c = 9</math>. Thus, <math>a < b < c</math>. |
| − | So, we have w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44. This means a + b + 9 is a multiple of 4. Testing values of a and b, we find (a, b, c) = (1, 6, 9), (3, 4, 5), and (5, 6, 9) all satisfy this relation. The corresponding (w, x, y, z) sets are (15, 14, 9, 6), (15, 12, 11, 6), and (16, 11, 10, 7). The first set does not satisfy the given conditions, but the other two do. Thus, w = 15 and w = 16 are both possible solutions so the answer is 15 + 16 = 31. | + | So, we have <math>w + x + y + z = w + (w - a) + (w - b) + (w - 9) = 4w - a - b - 9 = 44</math>. This means <math>a + b + 9</math> is a multiple of <math>4</math>. Testing values of <math>a</math> and <math>b</math>, we find <math>(a, b, c) = (1, 6, 9), (3, 4, 5),</math> and <math>(5, 6, 9)</math> all satisfy this relation. The corresponding <math>(w, x, y, z)</math> sets are <math>(15, 14, 9, 6), (15, 12, 11, 6),</math> and <math>(16, 11, 10, 7)</math>. The first set does not satisfy the given conditions, but the other two do. Thus, <math>w = 15</math> and <math>w = 16</math> are both possible solutions so the answer is <math>15 + 16 = 31</math>. |
| + | |||
| + | ~~latexed by jk23541 | ||
== See Also== | == See Also== | ||
Revision as of 03:49, 14 February 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 for 
?
Solution 1
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.
![[asy] unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1); pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1); draw(Z1--W1); draw(Z4--W4); pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4}; dot(ps); label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",W1,N); label("$z$",Z4,N); label("$y$",Y4,N); label("$x$",X4,N); label("$w$",W4,N); label("$1$",(X1--Y1),N); label("$1$",(X4--Y4),N); label("$3$",(Y1--Z1),N); label("$3$",(W4--X4),N); label("$5$",(X1--W1),N); label("$5$",(Y4--Z4),N); [/asy]](http://latex.artofproblemsolving.com/c/e/1/ce15bff994b2e1387b4b56d1de02fde16bfbfc54.png)
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 
Solution 2
First, like Solution 1, we know that 
, because no sum could be smaller. Next, we find the sum of all the differences; since is in the positive part of a difference 3 times, and has no differences where it contributes as the negative part, the sum of the differences includes 
. Continuing in this way, we find that ![\[3w+x-y-3z=28 \ \text{(2)}\]](http://latex.artofproblemsolving.com/e/b/5/eb562ead08862aa5f73f847f7a7198f11d5aba32.png)
. Now, we can subtract 
from (2) to get 
. Also, adding (2) with 
gives 
, or 
. Subtracting (1) from this gives 
. Since we know 
and 
, we find that ![\[(w-z)+(x-y)=(w-y)+(x-z)=9+1=10\]](http://latex.artofproblemsolving.com/0/2/e/02e2ad0c5e7a55f91f3bbc7f781b8c4c97cb5266.png)
. This means that 
and 
must be 4 and 6, in some order. If 
, then subtracting this from (3) gives 
, so 
. This means that 
, so 
. Similarly, can also equal 
.
Now if you are in a rush, you would have just answered 
. But we do have to check if these work. In fact, they do, giving solutions 
and 
.
Solution 3
Let 
, 
, 
. As above, we know that 
. Thus, 
.
So, we have 
. This means 
is a multiple of 
. Testing values of 
and 
, we find 
and 
all satisfy this relation. The corresponding 
sets are 
and 
. The first set does not satisfy the given conditions, but the other two do. Thus, 
and 
are both possible solutions so the answer is 
.
~~latexed by jk23541
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
