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8
2017 Team #8: floor(alpha^n) is divisible by 2017
2017 Team #8: floor(alpha^n) is divisible by 2017
Source:
February 19, 2017
floor function
Problem Statement
Does there exist an irrational number
α
>
1
\alpha > 1
α
>
1
such that
⌊
α
n
⌋
≡
0
(
m
o
d
2017
)
\lfloor \alpha^n \rfloor \equiv 0 \pmod{2017}
⌊
α
n
⌋
≡
0
(
mod
2017
)
for all integers
n
≥
1
n \ge 1
n
≥
1
?
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