# Repeated digit multiplication: How often does it lead to zero?

Take a number $n$ expressed in base $10$, and multiply its digits together to yield a new number. Repeat this process until a single digit is reached: \begin{eqnarray} 42433 &\to& 4 \cdot 2 \cdot 4 \cdot 3 \cdot 3 = 288 \\ 288 &\to& 2 \cdot 8 \cdot 8 = 128 \\ 128 &\to& 1 \cdot 2 \cdot 8 = 16 \\ 16 &\to& 6 \end{eqnarray}

Often a zero digit occurs along the way, which then maps to $0$ immediately: \begin{eqnarray} 31378 &\to& 3 \cdot 1 \cdot 3 \cdot 7 \cdot 8 = 504 \\ 504 &\to& 5 \cdot 0 \cdot 4 = 0 \end{eqnarray}

My question is:

Q. For a random number $n \in [1,N]$, what is the probability that this repeated digit multiplication leads to $0$?

It seems quite high, perhaps $90$% for large $N$, but I don't see a straightforward way to calculate it. It would also be of interest to know the distribution of the non-zero terminal digits. For example, even terminal digits occur more frequently than odd digits (for obvious reasons).

One can ask the same questions for numbers expressed in base $b$.

• if 0 is in the digits 100% same with 5 and any even number as the next value will then have a 0 in it ... – user451844 Aug 11 '17 at 0:59
• For large enough numbers of digits, the probability of getting a multiple of 10 after just one step approaches 1. – Daniel Schepler Aug 11 '17 at 0:59
• Thanks to Robert & Tony. I see now it is easy to establish that the probability approaches $1$, but perhaps difficult to compute it exactly for a given $N$. – Joseph O'Rourke Aug 11 '17 at 1:18

## 2 Answers

Given $\epsilon > 0$, for large enough $N$, more than $1-\epsilon$ of the integers in $[1,N]$ will have at least one $0$. That is, the fraction of $d$-digit integers with no $0$ is $(9/10)^{d-1}$, which goes to $0$ as $d \to \infty$.

• Also worth noting: This answer is essentially unchanged if we work in a different base. The fraction $9/10$ simply becomes $(b-1)/b$, and everything else stays the same. – G Tony Jacobs Aug 11 '17 at 1:09

The answer is greater than or equal to the proportion of numbers in $[1,N]$ that have the digit $0$, OR that have at least one $5$ and at least one even digit. As $N\to\infty$, this proportion approaches $1$.

The probability sought is greater than the proportion indicated above, because of numbers such as $2333$, whose digit product is $54$, which in turn produces a $0$ in the next step.