How Long Can a Number Resist?

Write a non-negative whole number in base 10 and multiply its digits together. If the product still has more than one digit, repeat with that product. The number of rounds needed to reach one digit is its multiplicative persistence.

For 39, calculate 3 × 9 = 27, then 2 × 7 = 14, then 1 × 4 = 4. This gives three rounds. A single-digit start already meets the stopping condition and needs zero rounds; a multi-digit start containing a zero reaches zero in one round.

Every start eventually finishes because a multi-digit number is larger than the product of its digits. That does not give a uniform upper bound on the number of rounds for arbitrarily long inputs.

The record examples let you inspect five rounds from 679, seven from 68,889 and eleven from 277,777,788,888,899. Whether any base-10 start needs twelve or more rounds remains an open question.

Two forces pull against each other, and watching them is most of the intuition.

Multiplying digits is a shrinking operation. Nine digits, each at most 9, multiply to at most 9⁹, which is nine digits' worth of number squeezed into nine digits of product at the very best and usually far fewer. A number and its digit product are not close.

But shrinking is not the same as shrinking fast. A digit 1 costs nothing at all: it leaves the product untouched. Small digits are nearly as cheap. So a number that wants to survive many rounds wants large digits, many of them, and above all it must contain no 0, because a single 0 collapses the whole product to 0 in one round and the routine is over.

That is why the record holders look the way they do. Run the sweep and read them: 679, 6788, 68889, 2677889. Sevens, eights and nines, no zeros, no ones, and each rung about ten times the size of the one before. Every extra round seems to cost roughly one more digit. Nobody has been able to turn that observation into an argument in either direction.

Here is the sentence this page exists to make impossible to misread.

That the routine always finishes is proved. How long it can take is completely open. Those are different questions, and settling the first says nothing whatever about the second.

The proof of termination is three lines and appears in the next layer but one. It works by showing each round strictly decreases the value, and a strictly decreasing sequence of non-negative integers has to stop. Notice what that argument never mentions: how much smaller. It would be equally happy with a routine that shed one unit per round and took a hundred trillion of them.

So the natural next question, "how many rounds can a number need", is untouched by the proof, and after fifty years it remains untouched by anything else. Published searches have gone to 10¹⁵ and found nothing needing more than eleven rounds. This page does not repeat that search; it uses two small lemmas to reach the fifteen-digit record holder directly, and cites the larger search for what it is.

A search that finds nothing is not an argument that there is nothing. It is a description of where somebody looked.

Fix a base $b\ge 2$. For $n$ written in base $b$ with digits $d_1{d}_2\dots d_k$, let

The multiplicative persistence ${\pi }_b(n)$ is the least $m\ge 0$ with $P_b^m(n)<b$, that is, the number of rounds needed to reach a single digit.

1. ${\pi }_b(n)$ is finite for every $n\ge 0$ and every $b\ge 2$. Proved.
2. ${\pi }_2(n)\le 1$ for every $n$. Proved.
3. $\min \{n:{\pi }_{10}(n)=11\}=277,777,788,888,899$, and ${\pi }_{10}(n)\le 11$ for every $n$ with at most 15 digits. Exhaustively checked over the stated range.
4. Whether there exists $n$ with ${\pi }_{10}(n)\ge 12$ is open.

Statement 1 is the one that makes the page a page: a routine that provably terminates, with no known bound at all on how long termination takes.

The routine terminates. Let $n$ have base-$b$ digits $d_1{d}_2\dots d_k$ with $k\ge 2$ and $d_1\ge 1$. Then

because each of $d_2,\dots ,d_k$ is at most $b-1$. So every round applied to a number of two or more digits produces a strictly smaller non-negative integer, and no strictly decreasing sequence of non-negative integers is infinite. The routine therefore reaches a single digit after finitely many rounds. ∎

In base 2 it finishes in one round. Every binary digit is 0 or 1. A number with $k\ge 2$ binary digits either contains a 0, giving $P_2(n)=0$, or is written entirely with 1s, giving $P_2(n)=1$. Both are single digits, so ${\pi }_2(n)\le 1$. ∎

The search for record holders can be cut down. Two observations, both used by this page and both cheap. First, $P_b$ depends only on the multiset of digits, and among all arrangements of a fixed multiset the smallest number is the one whose digits ascend; so only non-decreasing digit strings need testing. Second, if $n$ has $k\ge 3$ digits one of which is 1, deleting that 1 gives a smaller number with the same digit product and hence the same persistence, and any $n$ containing a 0 has ${\pi }_b(n)=1$; so a smallest example of persistence 2 or more contains no 0 and no 1. That reduces the fifteen-digit search from $10^{15}$ candidates to a few hundred thousand. ∎

What is not proved, and why the gap is not technical. The first proof gives no bound on ${\pi }_{10}$. To show that no number has persistence 12 you would have to constrain infinitely many digit strings at once, and the only structure available is that the product of the digits has to land, round after round, on another number with no 0 among its digits. Nobody has found a way to turn that into a bound, and nobody has found a number that needs twelve rounds.

What this laboratory computes. Runs are exact at every width, in arbitrary-precision integers, because the record holder is fifteen digits wide and the natural thing to type after seeing it is something wider. The sweep counts rounds for every start up to a hundred thousand. The record ladder is not swept at all: it is found by enumerating non-decreasing digit strings, which is the only reason a browser can reach fifteen digits.

What this laboratory does not compute. The published bound of 10¹⁵ is cited, not reproduced. This page rules out persistence 12 among fifteen-digit numbers whose digits ascend; ruling it out among all numbers below 10¹⁵ is a much larger computation, and it is somebody else's. The claim ledger keeps those two apart deliberately, and if you read only one line of it, read that one.

Read this next to happy numbers. The rules are near neighbours, squares of digits against a plain product of them, and the mathematics could hardly be more different. Happy numbers cannot escape a finite region, so enumerating that region settles every question about them and the answers are proofs. Multiplicative persistence has no finite region: the values fall, but the starts run off to infinity and no computation can follow them.

That is the useful thing to leave with. A proof that a process ends and a bound on how long it takes are separate results, and having the first is no evidence at all about the second. Fifty years of searching has produced eleven rounds and no argument, which is exactly what "open" means.