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Sweden Contests
Swedish Mathematical Competition
2011 Swedish Mathematical Competition
6
6
Part of
2011 Swedish Mathematical Competition
Problems
(1)
no of functions f (max {x + y + 2, xy} ) = min {f (x + y), f (xy + 2)}
Source: 2011 Swedish Mathematical Competition p6
4/30/2021
How many functions
f
:
N
→
N
f:\mathbb N \to \mathbb N
f
:
N
→
N
are there such that
f
(
0
)
=
2011
f(0)=2011
f
(
0
)
=
2011
,
f
(
1
)
=
111
f(1) = 111
f
(
1
)
=
111
, and
f
(
max
{
x
+
y
+
2
,
x
y
}
)
=
min
{
f
(
x
+
y
)
,
f
(
x
y
+
2
)
}
f\left(\max \{x + y + 2, xy\}\right) = \min \{f (x + y), f (xy + 2)\}
f
(
max
{
x
+
y
+
2
,
x
y
}
)
=
min
{
f
(
x
+
y
)
,
f
(
x
y
+
2
)}
for all non-negative integers
x
x
x
,
y
y
y
?
algebra
functional
functional equation