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2018 PUMaC Algebra A
2018 PUMaC Algebra A
Part of
2018 Princeton University Math Competition
Subcontests
(8)
8
1
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2018 PUMaC Algebra A8
p
q
=
∑
n
=
1
∞
1
2
n
+
6
(
10
−
4
cos
2
(
π
n
24
)
)
(
1
−
(
−
1
)
n
)
−
3
cos
(
π
n
24
)
(
1
+
(
−
1
)
n
)
25
−
16
cos
2
(
π
n
24
)
\frac{p}{q} = \sum_{n = 1}^\infty \frac{1}{2^{n + 6}} \frac{(10 - 4\cos^2(\frac{\pi n}{24})) (1 - (-1)^n) - 3\cos(\frac{\pi n}{24}) (1 + (-1)^n)}{25 - 16\cos^2(\frac{\pi n}{24})}
q
p
=
n
=
1
∑
∞
2
n
+
6
1
25
−
16
cos
2
(
24
πn
)
(
10
−
4
cos
2
(
24
πn
))
(
1
−
(
−
1
)
n
)
−
3
cos
(
24
πn
)
(
1
+
(
−
1
)
n
)
where
p
p
p
and
q
q
q
are relatively prime positive integers. Find
p
+
q
p + q
p
+
q
.
7
1
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2018 PUMaC Algebra A7
Let the sequence
{
a
n
}
n
=
−
2
∞
\left \{ a_n \right \}_{n = -2}^\infty
{
a
n
}
n
=
−
2
∞
satisfy
a
−
1
=
a
−
2
=
0
,
a
0
=
1
a_{-1} = a_{-2} = 0, a_0 = 1
a
−
1
=
a
−
2
=
0
,
a
0
=
1
, and for all non-negative integers
n
n
n
,
n
2
=
∑
k
=
0
n
a
n
−
k
a
k
−
1
+
∑
k
=
0
n
a
n
−
k
a
k
−
2
n^2 = \sum_{k = 0}^n a_{n - k}a_{k - 1} + \sum_{k = 0}^n a_{n - k}a_{k - 2}
n
2
=
k
=
0
∑
n
a
n
−
k
a
k
−
1
+
k
=
0
∑
n
a
n
−
k
a
k
−
2
Given
a
2018
a_{2018}
a
2018
is rational, find the maximum integer
m
m
m
such that
2
m
2^m
2
m
divides the denominator of the reduced form of
a
2018
a_{2018}
a
2018
.
6
1
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2018 PUMaC Algebra A6
Let
a
,
b
,
c
a, b, c
a
,
b
,
c
be non-zero real numbers that satisfy
1
a
b
c
+
1
a
+
1
c
=
1
b
\frac{1}{abc} + \frac{1}{a} + \frac{1}{c} = \frac{1}{b}
ab
c
1
+
a
1
+
c
1
=
b
1
. The expression
4
a
2
+
1
+
4
b
2
+
1
+
7
c
2
+
1
\frac{4}{a^2 + 1} + \frac{4}{b^2 + 1} + \frac{7}{c^2 + 1}
a
2
+
1
4
+
b
2
+
1
4
+
c
2
+
1
7
has a maximum value
M
M
M
. Find the sum of the numerator and denominator of the reduced form of
M
M
M
.
5
1
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2018 PUMaC Algebra A5
For
k
∈
{
0
,
1
,
…
,
9
}
,
k \in \left \{ 0, 1, \ldots, 9 \right \},
k
∈
{
0
,
1
,
…
,
9
}
,
let
ϵ
k
∈
{
−
1
,
1
}
\epsilon_k \in \left \{-1, 1 \right \}
ϵ
k
∈
{
−
1
,
1
}
. If the minimum possible value of
∑
i
=
1
9
∑
j
=
0
i
−
1
ϵ
i
ϵ
j
2
i
+
j
\sum_{i = 1}^9 \sum_{j = 0}^{i -1} \epsilon_i \epsilon_j 2^{i + j}
∑
i
=
1
9
∑
j
=
0
i
−
1
ϵ
i
ϵ
j
2
i
+
j
is
m
m
m
, find
∣
m
∣
|m|
∣
m
∣
.
4
1
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2018 PUMaC Algebra A4
Suppose real numbers
a
,
b
,
c
,
d
a, b, c, d
a
,
b
,
c
,
d
satisfy
a
+
b
+
c
+
d
=
17
a + b + c + d = 17
a
+
b
+
c
+
d
=
17
and
a
b
+
b
c
+
c
d
+
d
a
=
46
ab + bc + cd + da = 46
ab
+
b
c
+
c
d
+
d
a
=
46
. If the minimum possible value of
a
2
+
b
2
+
c
2
+
d
2
a^2 + b^2 + c^2 + d^2
a
2
+
b
2
+
c
2
+
d
2
can be expressed as a rational number
p
q
\frac{p}{q}
q
p
in simplest form, find
p
+
q
p + q
p
+
q
.
3
1
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2018 PUMaC Algebra A3
Let
x
0
,
x
1
,
…
x_0, x_1, \ldots
x
0
,
x
1
,
…
be a sequence of real numbers such that
x
n
=
1
+
x
n
−
1
x
n
−
2
x_n = \frac{1 + x_{n -1}}{x_{n - 2}}
x
n
=
x
n
−
2
1
+
x
n
−
1
for
n
≥
2
n \geq 2
n
≥
2
. Find the number of ordered pairs of positive integers
(
x
0
,
x
1
)
(x_0, x_1)
(
x
0
,
x
1
)
such that the sequence gives
x
2018
=
1
1000
x_{2018} = \frac{1}{1000}
x
2018
=
1000
1
.
2
1
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2018 PUMaC Algebra A2
If
a
1
,
a
2
,
…
a_1, a_2, \ldots
a
1
,
a
2
,
…
is a sequence of real numbers such that for all
n
n
n
,
∑
k
=
1
n
a
k
(
k
n
)
2
=
1
,
\sum_{k = 1}^n a_k \left( \frac{k}{n} \right)^2 = 1,
k
=
1
∑
n
a
k
(
n
k
)
2
=
1
,
find the smallest
n
n
n
such that
a
n
<
1
2018
a_n < \frac{1}{2018}
a
n
<
2018
1
.
1
1
Hide problems
2018 PUMaC Algebra A1
Let
a
k
=
0.
0
…
0
⏞
k
−
1
0
′
s
1
0
…
0
⏞
k
−
1
0
′
s
1
a_k = 0.\overbrace{0 \ldots 0}^{k - 1 \: 0's} 1 \overbrace{0 \ldots 0}^{k - 1 \: 0's} 1
a
k
=
0.
0
…
0
k
−
1
0
′
s
1
0
…
0
k
−
1
0
′
s
1
The value of
∑
k
=
1
∞
a
k
\sum_{k = 1}^\infty a_k
∑
k
=
1
∞
a
k
can be expressed as a rational number
p
q
\frac{p}{q}
q
p
in simplest form. Find
p
+
q
p + q
p
+
q
.