Polydivisible number

In mathematics a polydivisible number is a number with digits abcde... that has the following properties :-

# Its first digit a is not 0.
# The number formed by its first two digits ab is a multiple of 2.
# The number formed by its first three digits abc is a multiple of 3.
# The number formed by its first four digits abcd is a multiple of 4.
# etc.

For example, 345654 is a six-digit polydivisible number, but 123456 is not, because 1234 is not a multiple of 4. Polydivisible numbers can be defined in any base- however, the numbers in this article are all in base 10, so permitted digits are 0 to 9.

The smallest base 10 polydivisible numbers with 1,2,3,4... etc. digits are

1, 10, 102, 1020, 10200, 102000, 1020005, 10200056, 102000564, 1020005640 OEIS|id=A078282

Background

Polydivisible numbers are a generalisation of the following well-known problem in recreational mathematics :-

: "Arrange the digits 1 to 9 in order so that the first two digits form a multiple of 2, the first three digits form a multiple of 3, the first four digits form a multiple of 4 etc. and finally the entire number is a multiple of 9."

The solution to the problem is a nine-digit polydivisible number with the additional condition that it contains the digits 1 to 9 exactly once each. There are2,492 nine-digit polydivisible numbers, but the only one that satisfies the additional condition is

:381654729

How many polydivisible numbers are there?

If "k" is a polydivisible number with "n"-1 digits, then it can be extended to create a polydivisible number with "n" digits if there is a number between 10"k" and 10"k"+9 that is divisible by "n". If "n" is less or equal to 10, then it is always possible to extend an "n"-1 digit polydivisible number to an "n"-digit polydivisible number in this way, and indeed there may be more than one possible extension. If "n" is greater than 10, it is not always possible to extend a polydivisible number in this way, and as "n" becomes larger, the chances of being able to extend a given polydivisible number become smaller.

On average, each polydivisible number with "n"-1 digits can be extended to a polydivisible number with "n" digits in 10/"n" different ways. This leads to the following estimate of the number of "n"-digit polydivisible numbers, which we will denote by "F(n)" :-

:$$F(n) approx frac{9 imes 10^{n-1{n!}

Summing over all values of n, this estimate suggests that the total number of polydivisible numbers will be approximately

:$$frac{9(e^{10}-1)}{10}approx 19823

In fact, this underestimates the actual number of polydivisible numbers by about 3%.

Counting polydivisible numbers

We can find the actual values of "F(n)" by counting the number of polydivisible numbers with a given length :-

Length "n"	F("n")	Estimate of F("n")		Length "n"	F("n")	Estimate of F("n")		Length "n"	F("n")	Estimate of F("n")
1	9	9		11	2225	2255		21	18	17
2	45	45		12	2041	1879		22	12	8
3	150	150		13	1575	1445		23	6	3
4	375	375		14	1132	1032		24	3	1
5	750	750		15	770	688		25	1	1
6	1200	1250		16	571	430
7	1713	1786		17	335	253
8	2227	2232		18	180	141
9	2492	2480		19	90	74
10	2492	2480		20	44	37

There are 20,456 polydivisible numbers altogether, and the longest polydivisible number, which has 25 digits, is :-

:360 852 885 036 840 078 603 672 5

Related problems

Other problems involving polydivisible numbers include :-

* Finding polydivisible numbers with additional restrictions on the digits - for example, the longest polydivisible number that only uses even digits is

:480 006 882 084 660 840 40

* Finding palindromic polydivisible numbers - for example, the longest palindromic polydivisible number is

:300 006 000 03

* Enumerating polydivisible numbers in other bases.

External links

* [http://jwilson.coe.uga.edu/emt725/Class/Lanier/Nine.Digit/nine.html The nine-digit problem and its solution]
* [http://www.filmshuren.nl/randomstuff/polydivisiblenumbers.txt A list of all 20,456 polydivisible numbers]