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Canada Contests
Canada National Olympiad
1990 Canada National Olympiad
5
5
Part of
1990 Canada National Olympiad
Problems
(1)
Show that 0 ≤ f(n+1) - f(n) ≤ 1 and find n s.t. f(n) = 1025
Source: Canada National Mathematical Olympiad 1990 - Problem 5
10/4/2011
The function
f
:
N
→
R
f : \mathbb N \to \mathbb R
f
:
N
→
R
satisfies
f
(
1
)
=
1
,
f
(
2
)
=
2
f(1) = 1, f(2) = 2
f
(
1
)
=
1
,
f
(
2
)
=
2
and
f
(
n
+
2
)
=
f
(
n
+
2
−
f
(
n
+
1
)
)
+
f
(
n
+
1
−
f
(
n
)
)
.
f (n+2) = f(n+2 - f(n+1) ) + f(n+1 - f(n) ).
f
(
n
+
2
)
=
f
(
n
+
2
−
f
(
n
+
1
))
+
f
(
n
+
1
−
f
(
n
))
.
Show that
0
≤
f
(
n
+
1
)
−
f
(
n
)
≤
1
0 \leq f(n+1) - f(n) \leq 1
0
≤
f
(
n
+
1
)
−
f
(
n
)
≤
1
. Find all
n
n
n
for which
f
(
n
)
=
1025
f(n) = 1025
f
(
n
)
=
1025
.
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