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MAA AMC
AMC 10
2016 AMC 10
3
Absolute Value Mania
Absolute Value Mania
Source: Problem #3 2016 AMC 10B and 12A #3
February 21, 2016
absolute value
AMC
AMC 10
AMC 10 B
2016 AMC 10B
2016 AMC 12B
Problem Statement
Let
x
=
−
2016
x=-2016
x
=
−
2016
. What is the value of
∣
∣
∣
x
∣
−
x
∣
−
∣
x
∣
∣
−
x
\left| \ \bigl \lvert { \ \lvert x\rvert -x }\bigr\rvert -|x|{\frac{}{}}^{}_{}\right|-x
∣
x
∣
−
x
−
∣
x
∣
−
x
?
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
−
2016
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
0
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
2016
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4032
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
6048
<span class='latex-bold'>(A)</span>\ -2016\qquad<span class='latex-bold'>(B)</span>\ 0\qquad<span class='latex-bold'>(C)</span>\ 2016\qquad<span class='latex-bold'>(D)</span>\ 4032\qquad<span class='latex-bold'>(E)</span>\ 6048
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
−
2016
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
0
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
2016
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4032
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
6048
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