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International Contests
Austrian-Polish
1991 Austrian-Polish Competition
9
9
Part of
1991 Austrian-Polish Competition
Problems
(1)
f: A \to A, f(k)\ne g(k) and f(f(f(k))= g(k)$ for k \in A
Source: Austrian Polish 1991 APMC
5/1/2020
For a positive integer
n
n
n
denote
A
=
{
1
,
2
,
.
.
.
,
n
}
A = \{1,2,..., n\}
A
=
{
1
,
2
,
...
,
n
}
. Suppose that
g
:
A
→
A
g : A\to A
g
:
A
→
A
is a fixed function with
g
(
k
)
≠
k
g(k) \ne k
g
(
k
)
=
k
and
g
(
g
(
k
)
)
=
k
g(g(k)) = k
g
(
g
(
k
))
=
k
for
k
∈
A
k \in A
k
∈
A
. How many functions
f
:
A
→
A
f: A \to A
f
:
A
→
A
are there such that
f
(
k
)
≠
g
(
k
)
f(k)\ne g(k)
f
(
k
)
=
g
(
k
)
and
f
(
f
(
f
(
k
)
)
=
g
(
k
)
f(f(f(k))= g(k)
f
(
f
(
f
(
k
))
=
g
(
k
)
for
k
∈
A
k \in A
k
∈
A
?
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functional
composition
algebra