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1997 Nordic
4
4
Part of
1997 Nordic
Problems
(1)
f(2x) = 2f(x), f(4x+1) = 4f(x) + 3, f(4x-1) = 2f(2x -1)-1
Source: Nordic Mathematical Contest 1997 #4
10/4/2017
Let f be a function defined in the set
{
0
,
1
,
2
,
.
.
.
}
\{0, 1, 2,...\}
{
0
,
1
,
2
,
...
}
of non-negative integers, satisfying
f
(
2
x
)
=
2
f
(
x
)
,
f
(
4
x
+
1
)
=
4
f
(
x
)
+
3
f(2x) = 2f(x), f(4x+1) = 4f(x) + 3
f
(
2
x
)
=
2
f
(
x
)
,
f
(
4
x
+
1
)
=
4
f
(
x
)
+
3
, and
f
(
4
x
−
1
)
=
2
f
(
2
x
−
1
)
−
1
f(4x-1) = 2f(2x - 1) -1
f
(
4
x
−
1
)
=
2
f
(
2
x
−
1
)
−
1
. Show that
f
f
f
is an injection, i.e. if
f
(
x
)
=
f
(
y
)
f(x) = f(y)
f
(
x
)
=
f
(
y
)
, then
x
=
y
x = y
x
=
y
.
algebra
injective function
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