Let S(n) be the sum of the digits of a natural number n, e.g. S(168) = 15.
Call a number n "special" if it is evenly divisible by S(n), e.g. 36 is special because S(36) = 9 and 9 divides 36.
Prove that there cannot be a sequence of 22 consecutive numbers that are all special.
Bonus: Prove that there cannot be a sequence of 21 consecutive special numbers.
An even number can't divide an odd. This, for an odd number to be potentially special it must have an odd sod.
Within any set of 10 ten consecutive numbers of the form 10n + [0,1,...9] (call this a decade) the sod alternates even/odd. Thus some of these decades have don't have two potential special numbers in a row (the wrong kind). However, ever other decade has 10 numbers that could be special (the right kind).
These decades themselves alternate between the two types except at the multiples of 100 (centuries). So there can be consecutive two decades of the right kind they will look like
x90, x91, ... , x99, y00, y01, ... y09.
Where x99 has an odd sum (x has odd digit sum) and y01 also has an odd digit sum (y has even, implying x doesn't end in an odd number of 9's)
This is 20 potential special numbers. Since x89 has even sod it cant be special. y10 could be special, y11 has even sum so can't be special.
So there are sets of 21 potential special numbers but not 22.
Note: to prove the bonus, I would try to show that one endpoint rules out the other.
Edited on June 11, 2021, 11:14 am
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Posted by Jer
on 2021-06-11 11:05:35 |
Solution, not bonus
