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2000 Poland - Second Round
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5
Part of
2000 Poland - Second Round
Problems
(1)
Decide whether function exists
Source: 51 Polish MO 2000 Second Round - Problem 5
4/23/2018
Decide whether exists function
f
:
N
→
N
f: \mathbb{N} \rightarrow \mathbb{N}
f
:
N
→
N
, such that for each
n
∈
N
n \in \mathbb{N}
n
∈
N
is
f
(
f
(
n
)
)
=
2
n
f(f(n) )= 2n
f
(
f
(
n
))
=
2
n
.
algebra
number theory
function
Poland