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Champagnat numbers (Posted on 2018-01-23) Difficulty: 4 of 5
A Champagnat number is equal to the sum of all the digits in a set of consecutive positive integers, one of which is the number itself.
Thus, 42 is a Champagnat number, since 42 is the sum of all of the digits of 39, 40, 41, 42, 43, 44.

Prove that there exist infinitely many Champagnat numbers.

No Solution Yet Submitted by Ady TZIDON    
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re: a proof Comment 2 of 2 |
(In reply to a proof by Charlie)

My guess is the list of non-champagnat numbers is finite.

Not only do they dwindle, but they dwindle fast once you add a couple digits.  
Consider 5 digit numbers.  Each has a max sod of 9*5=45 so a very weak upper bound is 10000/45=222.22 numbers in any string.  Each string then contains at least 4 champ numbers (222/45).  Again this is a very weak upper bound.  
But if every starting number has to give us 4 champ numbers it would be very unlikely for a number to get missed every time.

For 6 digits the same reasoning gives 100000/54^2=34.29 champ numbers for each string.  It would be awful hard for a potential champ to slip through.

  Posted by Jer on 2018-01-23 16:01:25
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