# s69 - Solution to Problem 69 Congratulations to this week's...

This preview shows pages 1–2. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

s69 - Solution to Problem 69 Congratulations to this week's...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online