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2016 AIME Problems
6
p-addict
p-addict
Source: 2016 AIME II #6
March 17, 2016
AMC
AIME
AIME II
algebra
polynomial
Problem Statement
For polynomial
P
(
x
)
=
1
−
1
3
x
+
1
6
x
2
P(x)=1-\frac{1}{3}x+\frac{1}{6}x^2
P
(
x
)
=
1
−
3
1
x
+
6
1
x
2
, define
Q
(
x
)
=
P
(
x
)
P
(
x
3
)
P
(
x
5
)
P
(
x
7
)
P
(
x
9
)
=
∑
i
=
0
50
a
i
x
i
.
Q(x) = P(x)P(x^3)P(x^5)P(x^7)P(x^9) = \sum\limits_{i=0}^{50}a_ix^i.
Q
(
x
)
=
P
(
x
)
P
(
x
3
)
P
(
x
5
)
P
(
x
7
)
P
(
x
9
)
=
i
=
0
∑
50
a
i
x
i
.
Then
∑
i
=
0
50
∣
a
i
∣
=
m
n
\sum\limits_{i=0}^{50}|a_i|=\frac{m}{n}
i
=
0
∑
50
∣
a
i
∣
=
n
m
, where
m
m
m
and
n
n
n
are relatively prime positive integers. Find
m
+
n
m+n
m
+
n
.
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