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2001 Korea Junior Math Olympiad
7
2001 KJMO P7 (arrangement difference sum = n^2)
2001 KJMO P7 (arrangement difference sum = n^2)
Source: 2001 KJMO P7
June 29, 2024
algebra
combinatorics
set
Problem Statement
Finite set
{
a
1
,
a
2
,
.
.
.
,
a
n
,
b
1
,
b
2
,
.
.
.
,
b
n
}
=
{
1
,
2
,
…
,
2
n
}
\{a_1, a_2, ..., a_n, b_1, b_2, ..., b_n\}=\{1, 2, …, 2n\}
{
a
1
,
a
2
,
...
,
a
n
,
b
1
,
b
2
,
...
,
b
n
}
=
{
1
,
2
,
…
,
2
n
}
is given. If
a
1
<
a
2
<
.
.
.
<
a
n
a_1<a_2<...<a_n
a
1
<
a
2
<
...
<
a
n
and
b
1
>
b
2
>
.
.
.
>
b
n
b_1>b_2>...>b_n
b
1
>
b
2
>
...
>
b
n
, show that
∑
i
=
1
n
∣
a
i
−
b
i
∣
=
n
2
\sum_{i=1}^n |a_i-b_i|=n^2
i
=
1
∑
n
∣
a
i
−
b
i
∣
=
n
2
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