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Ireland Contests
Ireland National Math Olympiad
1991 Irish Math Olympiad
3
3
Part of
1991 Irish Math Olympiad
Problems
(1)
Constructing all Integers with 3 Functions
Source: 1991 IrMO Paper 1 Problem 3
10/1/2017
Three operations
f
,
g
f,g
f
,
g
and
h
h
h
are defined on subsets of the natural numbers
N
\mathbb{N}
N
as follows:
f
(
n
)
=
10
n
f(n)=10n
f
(
n
)
=
10
n
, if
n
n
n
is a positive integer;
g
(
n
)
=
10
n
+
4
g(n)=10n+4
g
(
n
)
=
10
n
+
4
, if
n
n
n
is a positive integer;
h
(
n
)
=
n
2
h(n)=\frac{n}{2}
h
(
n
)
=
2
n
, if
n
n
n
is an even positive integer. Prove that, starting from
4
4
4
, every natural number can be constructed by performing a finite number of operations
f
f
f
,
g
g
g
and
h
h
h
in some order.
[
[
[
For example:
35
=
h
(
f
(
h
(
g
(
h
(
h
(
4
)
)
)
)
)
)
.
]
35=h(f(h(g(h(h(4)))))).]
35
=
h
(
f
(
h
(
g
(
h
(
h
(
4
))))))
.
]
function
algebra