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Solution to Problem 69
Congratulations to this week’s winner
Nathan Pauli
Partial solutions were also received from Joshua Durham, Ray Kremer.
Additional partial solutions
were submitted by Jure Velkavrh and Ivan Lisac, Maxim Ovsjanikov, Philippe Fondanaiche, Burkart
Venzke.
Let’s start with a sequence
n
1
,
n
2
,...,
n
k
which can be moved to all zeroes.
In each
move, one of the entries is made zero; I’ll say that entry was
cleared
by the move.
It’s easy to see that the following "rules" must be followed:
1.
each entry
n
i
must be no greater than
i
after each move,
2.
if there are two entries which may be cleared, it’s the smaller of the two which
must be cleared.
To see what the value of each
n
i
should be, let
q
i
denote the total number of times
that the
i
’th entry is cleared.
The
i
’th entry started as
n
i
and was increased one time
for each time every time that a higher entry was cleared.
Hence [IMAGE] is the
number of times the
i
’th entry had the value
i
, i.e.
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 Spring '10
 AlbertoDelgado
 Combinatorics, Addition

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