# One-digit products in a row of numbers

The digits from 1 to 9 can be arranged in a row, such that any two neighbouring digits in this row is the product of two one-digit numbers. Arrangement:

$$728163549$$

Is it possible to do such an arrangement using hexadecimal digits 1-9 and A-F? Here the row has 15 digits and all numbers are treated as hexadecimal numbers. Example: 123456789ABCDEF 12=2*9, 23=5*7, 34=4*D, 45 does not work, etc.

• Neat! You could generalize to an OEIS entry of such numbers from base-3 upwards... in case of ambiguity, taking the numerically smallest. I wonder if there's a constructive proof that such a number exists for every base N. – smci Sep 23 at 16:59
• @smci The first base which has a solution is base $7$, with a single solution. Base $8$ also has a single solution while base $9$ oddly has three. – Jens Sep 23 at 18:09
• I posted a Generalization to base-N – smci Sep 23 at 20:20

One solution is

$$D2379A5B6C4E18F$$

Thought process:

No product starts with $$F$$, so $$F$$ must be at the end, and the only options are $$3F$$ and $$8F$$.
The only $$2$$-digit numbers that are products of $$1$$-digit number and start with digits $$A, B, C, D, E$$ are $$A5, A8, A9, B4, B6, C3, C4, D2, E1.$$
Therefore we must have subsequences $$E1$$ and $$D2$$. From this, you quickly get an answer by looking at the above table. I don't know if I was lucky, but apart from the observations above, I guessed all the rest right. Just for the sake of it, here is another one: $$D24E1879A5B6C3F$$

• It would be interesting to see if this can be generalized to all number bases. You can obviously play around with it like you did here, but proving it mathematically for all bases without having to test them individually might be an interesting challenge... – Darrel Hoffman Sep 23 at 13:52

The smallest such number is 375B6E19C4D2A8F. According to the brute-force program I made, there are just $$787$$ solutions.
• @ThomasL That's because as the amount of digits increases from 10 to 16, the amount of acceptable pairs increases roughly from $\frac{10^2}{2}$ to $\frac{16^2}{2}$ which is a huge increase in comparison. Statistically, there has to be a larger and larger number of solutions as the base increases. – Arnaud Mortier Sep 23 at 20:27