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Austria Contests
Austrian MO National Competition
2005 Federal Competition For Advanced Students, Part 2
1
f(2x+1)=f(2x), f(3x+1)=f(3x), f(5x+1)=f(5x)
f(2x+1)=f(2x), f(3x+1)=f(3x), f(5x+1)=f(5x)
Source: Austrian MO 2005 round 2
June 27, 2005
function
combinatorics unsolved
combinatorics
Problem Statement
The function
f
:
(
0
,
.
.
.
2005
)
→
N
f : (0,...2005) \rightarrow N
f
:
(
0
,
...2005
)
→
N
has the properties that
f
(
2
x
+
1
)
=
f
(
2
x
)
f(2x+1)=f(2x)
f
(
2
x
+
1
)
=
f
(
2
x
)
,
f
(
3
x
+
1
)
=
f
(
3
x
)
f(3x+1)=f(3x)
f
(
3
x
+
1
)
=
f
(
3
x
)
and
f
(
5
x
+
1
)
=
f
(
5
x
)
f(5x+1)=f(5x)
f
(
5
x
+
1
)
=
f
(
5
x
)
with
x
∈
(
0
,
1
,
2
,
.
.
.
,
2005
)
x \in (0,1,2,...,2005)
x
∈
(
0
,
1
,
2
,
...
,
2005
)
. How many different values can the function assume?
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