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Contests
National and Regional Contests
Switzerland Contests
Switzerland - Final Round
2019 Switzerland - Final Round
2
2
Part of
2019 Switzerland - Final Round
Problems
(1)
2019 Switzerland MO P2
Source: 2019 Switzerland MO
3/10/2019
Let
P
\mathbb{P}
P
be the set of all primes and let
M
M
M
be a subset of
P
\mathbb{P}
P
with at least three elements. Suppose that for all
k
≥
1
k \geq 1
k
≥
1
and for all subsets
A
=
{
p
1
,
p
2
,
…
,
p
k
}
A=\{p_1,p_2,\dots ,p_k \}
A
=
{
p
1
,
p
2
,
…
,
p
k
}
of
M
M
M
,
A
≠
M
A\neq M
A
=
M
, all prime factors of
p
1
p
2
…
p
k
−
1
p_1p_2\dots p_k-1
p
1
p
2
…
p
k
−
1
are in
M
M
M
. Prove that
M
=
P
M=\mathbb{P}
M
=
P
.
number theory