Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n(N)$?

share|cite|improve this question
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
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.

share|cite|improve this answer

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.

share|cite|improve this answer
Thank you for the response. – NikolajK Jan 15 '12 at 21:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.