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If I take any natural number, add the digits together until they produce a one digit number, like in the following example $$5847\ \ \rightarrow\ \ 5+8+4+7=24\ \ \rightarrow\ \ 2+4=6$$ which single digit number $n$ to end up with is the most common?

If the answer depends on the maximum natural number $N$, then what is the most probable $n(N)$?

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The result is the remainder of the original number on division by 9, which (for large $N$) is equally likely to be any of the digits 1, 2, ..., 9. –  Gerry Myerson Jan 15 '12 at 20:55
    
@Gerry, except that the remainder is never 9, and the iterated sum is (almost) never 0 ... –  Henning Makholm Jan 15 '12 at 21:02
    
@Henning, the remainder can be 9, if you want it to be. That is, if the number is a multiple of 9, then, for the purposes of this question only, define the remainder to be 9, and it all works. –  Gerry Myerson Jan 15 '12 at 21:07
    
@Gary: Sure, you can redefine things such that they work :-) –  Henning Makholm Jan 15 '12 at 21:09

2 Answers 2

up vote 1 down vote accepted

If your initial number $k$ is a positive integer, the iterated digit sum $n$ is simply the number between 1 and 9, inclusive, such that $k-n$ is a multiple of 9.

Therefore, if your maximum $N$ is a multiple of 9, then all of the 9 possible sums are equally likely. (I'm assuming that you're selecting $k$ uniformly between 1 and $N$, inclusive).

If $N$ is not a multiple of $9$, say $N=9S+a$ then results from $1$ through $a$ will be slightly more probable than results above $a$ -- but the larger $N$ becomes, the slighter will this difference be.

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The number you end with is the remainder of the original number upon division by $9$. For example, $5847 = 649 \cdot 9 + 6$. So far large $N$, all results $1-9$ will be roughly equiprobable. You could also calculate the exact distribution, if you really wanted.

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Thank you for the response. –  NikolajK Jan 15 '12 at 21:07

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