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freemind
Riemann Hypothesis
Riemann Hypothesis


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#1
Again Warm-Up Prob for a TST-Partition of a set into triples
Moldova 2008 IMO-BMO First TST Problem 2

We say the set \{1,2,\ldots,3k\} has property D if it can be partitioned into disjoint triples so that in each of them a number equals the sum of the other two.

(a) Prove that \{1,2,\ldots,3324\} has property D.

(b) Prove that \{1,2,\ldots,3309\} hasn't property D.
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The fate of equilibrium is to end the eternity...

PostPosted: Mon Mar 03, 2008 5:58 pm
Xell
P versus NP
P versus NP

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#2
(a) suppose that if D_{n}=\{1,2,...,n\} has property D, then so have D_{4n} and D_{4n+3} :

suppose D_{n} has property D.
D_{4n} can be partitioned into the following sets (in which the last integer is the sum of the two first :
\{1,2n+n,2n+n+1\}
\{3,2n+n-1,2n+n+2\}
\{5,2n+n-2,2n+n+3\}
...
\{2n-1,2n+1,4n\}
and eventually the even integers : \{2,4,6,...,2n\}=2D_{n}
if D_{n} has property D, 2D_{n} also has property D and hence D_{4n} has property D

D_{4n+3} can be partitioned in a similar way :
\{1,2n+n+2,2n+n+3\}
\{3,2n+n+1,2n+n+4\}
\{5,2n+n,2n+n+5\}
...
\{2n+1,2n+2,4n+3\}
and the even integers : \{2,4,6,...,2n\}=2D_{n}

so we have proved that if D_{n} has property D, so have D_{4n} and D_{4n+3}
now it is sufficient to check that :
- D_{3} has property D
- 3x4=12
- 12x4+3=51
- 51x4+3=207
- 207x4+3=831
- 831x4=3324
From that we conclude D_{3324} has property D.

(b) It is clear that D_{n} cannot have property D if the sum of its elements is odd : suppose D_{n} has property D ; by construction, the sum of the elements of each triple in the partition is even so the same must be true of D_{n}. So D_{3309} cannot have property D because the sum of its elements is an odd integer.

PostPosted: Thu Mar 06, 2008 1:07 am
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