Difference between revisions of "2011 AMC 10B Problems/Problem 21"
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== Solution 4 == | == Solution 4 == | ||
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== See Also== | == See Also== | ||
{{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}} | {{AMC10 box|year=2011|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:45, 26 March 2018
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
. 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
. 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
.
~~ solution by ccx09 Roy Short ~~latexed by jk23541
Solution 4
From
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 |
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