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Serbia National Math Olympiad
2013 Serbia National Math Olympiad
1
1
Part of
2013 Serbia National Math Olympiad
Problems
(1)
Bijection f satisfying |i-j| <= k -> |f(i) - f(j)|<=k
Source: Serbian National Olympiad 2013, Problem 1
4/8/2013
Let
k
k
k
be a natural number. Bijection
f
:
Z
→
Z
f:\mathbb{Z} \rightarrow \mathbb{Z}
f
:
Z
→
Z
has the following property: for any integers
i
i
i
and
j
j
j
,
∣
i
−
j
∣
≤
k
|i-j|\leq k
∣
i
−
j
∣
≤
k
implies
∣
f
(
i
)
−
f
(
j
)
∣
≤
k
|f(i) - f(j)|\leq k
∣
f
(
i
)
−
f
(
j
)
∣
≤
k
. Prove that for every
i
,
j
∈
Z
i,j\in \mathbb{Z}
i
,
j
∈
Z
it stands:
∣
f
(
i
)
−
f
(
j
)
∣
=
∣
i
−
j
∣
.
|f(i)-f(j)|= |i-j|.
∣
f
(
i
)
−
f
(
j
)
∣
=
∣
i
−
j
∣.
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