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In basic math, when adding there is the notion of carry. For example if you have $9999+9999$, since $9+9=18$ the $1$ is leftover and carried to the top. Is there some formula that gives us the total number carried? So in the case of our example, it would be $4$.

user327811
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  • In $a+b$, it can be found by $|S(a)+S(b)-S(a+b)|$ which can express the total number of carries in that expression. $S(n)$ is the sum of digits for $n$. – S.C.B. Apr 01 '16 at 18:33
  • In the sum of $a=11$ and $b=29$, the number of carries is $1$ but $S(a)+S(b)-S(a+b)=(1+1)+(2+9)-(4+0)=2+11-4=9$. Am I missing something? – Darío G Apr 01 '16 at 18:37
  • @MXYMXY Yes, this doesn't really make sense – user327811 Apr 01 '16 at 18:38
  • Obviously, I'm not correctly remembering the formula. But there was a connection. I just don't remember what. – S.C.B. Apr 01 '16 at 18:39
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    Every time you carry, you reduce the sum of the digits by 9. Take MXMXY's formula and divide by 9. – Doug M Apr 01 '16 at 18:43
  • @DougM Would you add that as an answer? – S.C.B. Apr 01 '16 at 18:47
  • I think MXMXY/DougM observation works in any base: the result will always be $\dfrac{S(a)+S(b)-S(a+b)}{n-1}$ if we are working in base $n$. – Darío G Apr 01 '16 at 19:03

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Kummer's theorem says the number of carries when adding $m + n$ in a prime base $p$ is the largest integer $k$ such that $p^k$ divides the binomial coefficient ${m + n \choose m}$. Unfortunately, $10$ is not a prime.

Robert Israel
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