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2024 CMIMC
2024 CMIMC Algebra and Number Theory
4
4
Part of
2024 CMIMC Algebra and Number Theory
Problems
(1)
2024 Alg/NT Problem 4
Source:
4/14/2024
For positive integer
n
n
n
, let
f
(
n
)
f(n)
f
(
n
)
be the largest integer
k
k
k
such that
k
!
≤
n
k!\leq n
k
!
≤
n
, let
g
(
n
)
=
n
−
(
f
(
n
)
)
!
g(n)=n-(f(n))!
g
(
n
)
=
n
−
(
f
(
n
))!
, and for
j
≥
1
j\geq 1
j
≥
1
let
g
j
(
n
)
=
g
(
…
(
g
(
n
)
)
…
)
⏟
j
times
.
g^j(n)=\underbrace{g(\dots(g(n))\dots)}_{\text{$j$ times}}.
g
j
(
n
)
=
j
times
g
(
…
(
g
(
n
))
…
)
.
Find the smallest positive integer
n
n
n
such that
g
j
(
n
)
>
0
g^{j}(n)> 0
g
j
(
n
)
>
0
for all
j
<
30
j<30
j
<
30
and
g
30
(
n
)
=
0
g^{30}(n)=0
g
30
(
n
)
=
0
.Proposed by Connor Gordon
number theory