# Sum a number's digits until only one digit remain

I want to iteratively calculate the sum of a number's digits, until the result contains only one digit. For example, when $$n=67946$$ we get:

$$6+7+9+4+6=32\space\to\space3+2=5\tag1$$

So, when $$n=67946$$ I must get $$5$$.

When $$n=649134976$$

$$6+4+9+1+3+4+9+7+6=49\space\to\space4+9=13\space\to\space1+3=4\tag2$$

So, when $$n=649134976$$ I must get $$4$$.

I wrote:

Total[IntegerDigits[n]]


But this only gives the first iteration of the sum of digits.

• @infinitezero I stand corrected (and deleted the comment). Thanks! I think Mod[n,9,1] does the trick. Dec 27, 2021 at 3:19

Clear[s, t, n]

t[n_] := NestWhileList[Total[IntegerDigits[#]] & , n, # > 10 &]


Test:

t[649134976]


{649134976, 49, 13, 4}

If you don't want the interim values, use NestWhile.

A slight variation:

s[n_] := FixedPointList[Total[IntegerDigits[#]] & , n]


Test:

s[649134976]


{649134976, 49, 13, 4, 4}

If you don't want the interim values, use FixedPoint.

• Nice solution. I think the FixedPoint version is a bit neater
– Jojo
Dec 26, 2021 at 19:23