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2
2017 Algebra/NT #2: Geometric Series
2017 Algebra/NT #2: Geometric Series
Source:
February 19, 2017
geometric series
Problem Statement
Find the value of
∑
1
≤
a
<
b
<
c
1
2
a
3
b
5
c
\sum_{1\le a<b<c} \frac{1}{2^a3^b5^c}
1
≤
a
<
b
<
c
∑
2
a
3
b
5
c
1
(i.e. the sum of
1
2
a
3
b
5
c
\frac{1}{2^a3^b5^c}
2
a
3
b
5
c
1
over all triples of positive integers
(
a
,
b
,
c
)
(a, b, c)
(
a
,
b
,
c
)
satisfying
a
<
b
<
c
a<b<c
a
<
b
<
c
)
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