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International Mathematical Arhimede Contest (IMAC)
2014 IMAC Arhimede
1
f(2)=0, f(3)> 0, f(6042)=2014, f(m+n)- f(m)-f(n) \in {0,1}
f(2)=0, f(3)> 0, f(6042)=2014, f(m+n)- f(m)-f(n) \in {0,1}
Source: IMAC Arhimede 2014 p1
May 6, 2019
functional equation
functional equation in N
algebra
Problem Statement
The function
f
:
N
→
N
0
f: N \to N_0
f
:
N
→
N
0
is such that
f
(
2
)
=
0
,
f
(
3
)
>
0
,
f
(
6042
)
=
2014
f (2) = 0, f (3)> 0, f (6042) = 2014
f
(
2
)
=
0
,
f
(
3
)
>
0
,
f
(
6042
)
=
2014
and
f
(
m
+
n
)
−
f
(
m
)
−
f
(
n
)
∈
{
0
,
1
}
f (m + n)- f (m) - f (n) \in\{0,1\}
f
(
m
+
n
)
−
f
(
m
)
−
f
(
n
)
∈
{
0
,
1
}
for all
m
,
n
∈
N
m,n \in N
m
,
n
∈
N
. Determine
f
(
2014
)
f (2014)
f
(
2014
)
.
N
0
=
{
0
,
1
,
2
,
.
.
.
}
N_0=\{0,1,2,...\}
N
0
=
{
0
,
1
,
2
,
...
}
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