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Czech-Polish-Slovak Match
2007 Czech-Polish-Slovak Match
5
5
Part of
2007 Czech-Polish-Slovak Match
Problems
(1)
{1,2,...,n} partitioned into triplets such that a+b=c
Source: Czech-Polish-Slovak Match 2007-P5
9/14/2011
For which
n
∈
{
3900
,
3901
,
⋯
,
3909
}
n\in\{3900, 3901,\cdots, 3909\}
n
∈
{
3900
,
3901
,
⋯
,
3909
}
can the set
{
1
,
2
,
.
.
.
,
n
}
\{1, 2, . . . , n\}
{
1
,
2
,
...
,
n
}
be partitioned into (disjoint) triples in such a way that in each triple one of the numbers equals the sum of the other two?
combinatorics proposed
combinatorics