(A)
16
(B)
31
(C)
48
(D)
62
(E)
93
2011 AMC 10B, Problem #21—
2011 AMC 12B, Problem #13—
“Write the largest pairwise difference using the variables.”
Solution
Answer (B):
The largest pairwise difference is 9, so
w
−
z
= 9
. Let
n
be either
x
or
y
. Because
n
is between
w
and
z
,
9 =
w
−
z
= (
w
−
n
) + (
n
−
z
)
.
Therefore the positive differences
w
−
n
and
n
−
z
must sum to 9. The given pairwise differences that sum to 9 are
3 + 6
and
4 + 5
. The remaining pairwise difference must be
x
−
y
= 1
.
The second largest pairwise difference is 6, so either
w
−
y
= 6
or
x
−
z
= 6
.
In the first case the set of four
numbers may be expressed as
{
w, w
−
5
, w
−
6
, w
−
9
}
. Hence
4
w
−
20 = 44
, so
w
= 16
. In the second case
w
−
x
= 3
, and the four numbers may be expressed as
{
w, w
−
3
, w
−
4
, w
−
9
}
. Therefore
4
w
−
16 = 44
, so
w
= 15
. The sum of the possible values for
w
is
16 + 15 = 31
.
Note: The possible sets of four numbers are
{
16
,
11
,
10
,
7
}
and
{
15
,
12
,
11
,
6
}
.
Diﬃculty:
Hard
CCSS-M:
A-CED.3.
Represent constraints by equations or inequalities, and by systems of equations
and/or inequalities, and interpret solutions as viable or nonviable options in a modeling context.
21

A pyramid has a square base with sides of length 1 and has
lateral faces that are equilateral triangles.
A cube is placed
within the pyramid so that one face is on the base of the
pyramid and its opposite face has all its edges on the lateral
faces of the pyramid. What is the volume of this cube?
(A)
5
√
2
−
7
(B)
7
−
4
√
3
(C)
2
√
2
27
(D)
√
2
9
(E)
√
3
9
2011 AMC 10B, Problem #22—
2011 AMC 12B, Problem #18—
“What is the height of the cube?”
Solution
Answer (A):
Let
A
be the apex of the pyramid, and let the base be the square
BCDE
. Then
AB
=
AD
= 1
and
BD
=
√
2
, so
△
BAD
is an isosceles right triangle. Let the cube have edge length
x
. The intersection of the cube
with the plane of
△
BAD
is a rectangle with height
x
and width
√
2
x
. It follows that
√
2 =
BD
= 2
x
+
√
2
x
,
from which
x
=
√
2
−
1
.
√
2
x
√
2
x
A
D
B
Hence the cube has volume
(
√
2
−
1)
3
= (
√
2)
3
−
3(
√
2)
2
+ 3
√
2
−
1 = 5
√
2
−
7
.
OR
Let
A
be the apex of the pyramid, let
O
be the center of the base, let
P
be the midpoint of one base edge, and let
the cube intersect
AP
at
Q
. Let a coordinate plane intersect the pyramid so that
O
is the origin,
A
on the positive
y
-axis, and
P
=
(
1
2
,
0
)
. Segment
AP
is an altitude of a lateral side of the pyramid, so
AP
=
√
3
2
, and it follows
that
A
=
(
0
,
√
2
2
)
. Thus the equation of line
AP
is
y
=
√
2
2
−
√
2
x
. If the side length of the cube is
s
, then
Q
= (
s
2
, s
)
, so
s
=
√
2
2
−
√
2
·
s
2
. Solving gives
s
=
√
2
−
1
, and the result follows that in the first solution.
Diﬃculty:
Hard
CCSS-M:
G-GMD.4. Identify the shapes of two-dimensional cross-sections of three-dimensional objects,
and identify three-dimensional objects generated by rotations of two-dimensional objects.
22

What is the hundreds digit of
2011
2011
?
(A)
1
(B)
4
(C)
5
(D)
6
(E)
9
2011 AMC 10B, Problem #23—
“Express as
(2000+11)
2011
and find all the terms divisible by 1000.”
Solution
Answer (D):
In the expansion of
(2000 + 11)
2011
, all terms except
11
2011
are divisible by 1000, so the hundreds
digit of
2011
2011
is equal to that of
11
2011
. Furthermore, in the expansion of
(10 + 1)
2011
, all terms except
1
2011
,
(
2011
1
)
(10)
(
1
2010
)
, and
(
2011
2
)
(10)
2
(
1
2009
)
are divisible by 1000. Thus the hundreds digit of
2011
2011
is equal
to that of
1 +
(
2011
1
)
(10)
(
1
2010
)
+
(
2011
2
)
(10)
2
(
1
2009
)
= 1 + 2011
·
10 + 2011
·
1005
·
100
= 1 + 2011
·
100510
.

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