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Poland Contests
Poland - Second Round
2019 Poland - Second Round
4
4
Part of
2019 Poland - Second Round
Problems
(1)
Coprime integers divisibility
Source: 2019 Second Round - Poland
7/8/2019
Let
a
1
,
a
2
,
…
,
a
n
a_1, a_2, \ldots, a_n
a
1
,
a
2
,
…
,
a
n
(
n
≥
3
n\ge 3
n
≥
3
) be positive integers such that
g
c
d
(
a
1
,
a
2
,
…
,
a
n
)
=
1
gcd(a_1, a_2, \ldots, a_n)=1
g
c
d
(
a
1
,
a
2
,
…
,
a
n
)
=
1
and for each
i
∈
{
1
,
2
,
…
,
n
}
i\in \lbrace 1,2,\ldots, n \rbrace
i
∈
{
1
,
2
,
…
,
n
}
we have
a
i
∣
a
1
+
a
2
+
…
+
a
n
a_i|a_1+a_2+\ldots+a_n
a
i
∣
a
1
+
a
2
+
…
+
a
n
. Prove that
a
1
a
2
…
a
n
∣
(
a
1
+
a
2
+
…
+
a
n
)
n
−
2
a_1a_2\ldots a_n | (a_1+a_2+\ldots+a_n)^{n-2}
a
1
a
2
…
a
n
∣
(
a
1
+
a
2
+
…
+
a
n
)
n
−
2
.
number theory
positive integers
Divisibility