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CentroAmerican
2006 CentroAmerican
3
3
Part of
2006 CentroAmerican
Problems
(1)
Function involving floors
Source: Central American Olympiad 2006, Problem 3
4/30/2007
For every natural number
n
n
n
we define
f
(
n
)
=
⌊
n
+
n
+
1
2
⌋
f(n)=\left\lfloor n+\sqrt{n}+\frac{1}{2}\right\rfloor
f
(
n
)
=
⌊
n
+
n
+
2
1
⌋
Show that for every integer
k
≥
1
k \geq 1
k
≥
1
the equation
f
(
f
(
n
)
)
−
f
(
n
)
=
k
f(f(n))-f(n)=k
f
(
f
(
n
))
−
f
(
n
)
=
k
has exactly
2
k
−
1
2k-1
2
k
−
1
solutions.
function
floor function
algebra proposed
algebra