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Most 'self-divisible' decimal digit number?
Consider the number x = 31250. It is divisible by any number in the following list containing 10 numbers: {1, 2, 5, 25, 50, 125, 250, 1250, 3125, 31250} which are all substrings of the decimal representation of x. The number 31250 has 14 divisors and you can form 15 substrings of it. Let the 'self-divisibility' of x be denoted by the product:
selfdiv(x) = (10/14)·(10/15) = 10/21 ≈ 47,619%
This far I have found the numbers 1, 15, 125, 1250 and 31250 as the numbers of greatest 'self-divisibility' when considering a fixed number of digits (they have 1, 2, 3, 4 and 5 digits repspectively). They are 100%, 75%, 66.7%, 64% and 47.6% 'self-divisible' respectively.
Surprisingly, for each fixed number of digits a unique number of maximal 'self-divisibility' came out in these five cases. I do not know whether this is the case for numbers having 6 digits or more.
Who can determine the number(s) that is most 'self-divisibile' among numbers of 6 and 7 digits. Can anyone go beyond 7 digits?
Samewise kindly pointed out the possibility of repeated substrings:
If x is an n-digit number and has m substrings that divides x out of the n·(n+1)/2 substrings you can form (some of them possibly identical) then subdiv(x) = 2m/(n²+n). If the substrings of x contain k out of a total of d distinct divisors of x, define:
selfdiv(x) = k/d·subdiv(x) = 2m·k/[d·(n²+n)]
With this 11 (not 15) becomes the most 'self-divisible' 2-digit number (being 100% 'self-divisible') and 125 shares the 3-digit highscore with 101 (and still 66.7%). Large values of selfdiv(x) corresponds to x's that has both a high percentage of it's substrings as divisors and a high percentage of it's divisors as substrings.
I meant Samwise, sorry! 'Samewise' just seemed so right speaking of 'repetitions'.
@Samwise: That's some extraordinary data you have produced! I can confirm your 6-digit winner - I found the same. I'll have to agree with you that both definitions has some disadvantages. The new repetition-adjusted one reflects my main idea about having a high percentage of both the divisors and the substrings at the same time.
That we are working with the decimal system does show, as your data reveals that numbers having 2 and 5 as multiple factors seem to have an advantage. If the formula were to be adjusted, I think it would be interesting to rank numbers having many unique substrings and many divisors over others.
Maybe an analytical approach could rule out many of the numbers needed to check through in accordance with the observation that 2 and 5 has to have a high representation?
2 Answers
- SamwiseLv 78 years agoFavorite Answer
While I appreciate the attempt to revise the formula to deal with repetitions of strings, and I perceive the motive behind the change, it turns out to have an unexpected result--one that I suspect shifts the preferences in an undesired direction.
My latest run of the program still has some characteristics I don't like, but it's time to report an excerpt from the output. I have edited out a lot: the program simply writes a report each time it finds a more self-divisible number (by either formula) for the digit-length it's working on. I'm omitting reports that were superseded. The fraction in brackets is the original formula, disregarding repeated substrings.
Start: Sun Jun 23 15:12:13 2013
1 (1 divisors): 1/1 [1/1]
factors:
substring divisors: 1
Up to 10: Sun Jun 23 15:12:13 2013 [4 primes found]
11 (2 divisors): 1/1 [2/3]
substring divisors: 11 1(2)
15 (4 divisors): 3/4 [3/4]
factors: 3 5
substring divisors: 1 15 5
Up to 100: Sun Jun 23 15:12:13 2013 [25 primes found]
101 (2 divisors): 2/3 [1/3]
factors:
substring divisors: 1(3) 101
125 (4 divisors): 2/3 [2/3]
factors: 5^3
substring divisors: 25 1 125 5
Up to 1000: Sun Jun 23 15:12:13 2013 [168 primes found]
1250 (10 divisors): 16/25 [16/25]
factors: 2 5^4
substring divisors: 25 50 250 2 1 1250 125 5
Up to 10000: Sun Jun 23 15:12:13 2013 [1229 primes found]
31250 (14 divisors): 10/21 [10/21]
factors: 2 5^6
substring divisors: 25 50 250 2 1 3125 1250 31250 125 5
Up to 100000: Sun Jun 23 15:12:40 2013 [9592 primes found]
100250 (16 divisors): 45/112 [27/112]
factors: 2 5^3 401
substring divisors: 25(3) 50 250(3) 2(3) 1 10025 10 100250 5
312500 (24 divisors): 7/18 [7/18]
factors: 2^2 5^7
substring divisors: 2 1 2500 31250 125 500 25 50 250 3125 1250 312500 12500 5
Up to 1000000: Sun Jun 23 15:38:18 2013 [78498 primes found]
The extra information was provided to guide me in improving the program. The timestamps are the most worrisome item: it took the computer about 58 times as long to scale up from 5 to 6 digits. I can shrink that by taking out some of the information-gathering, to bring that down from an unacceptable 25-hour run to obtain 7-digit results to a potentially acceptable overnight run.
Before I do so, we need to settle on the most desirable formula for comparing self-divisibility. The old formula picked 312500 as the best 6-digit number.
The new one, attempting to take into account repetitions of the same string, picked 100250. The reason is leading zeros: each of the substrings 2, 5, 25, 250 also appeared with one and two leading zeros. That's the reason each of those divisors got counted three times.
I could go back to ignoring repetitions, or use the formula with repetitions (which IMO overly favors numbers with early strings of zeros), or apply a new adapted formula. But I can't attempt a decent attack on the 7-digit range without restricting it to a single formula, unless we port the program from my computer to something more powerful, or more able to run the problem in background over a long time. The program is in Perl.
- az_lenderLv 78 years ago
You should possibly contribute this problem either to "Mathematics Magazine" or to the "American Mathematical Monthly." Both carry (or at least, used to carry) monthly sections of what by Yahoo! standards would be considered advanced questions. Your question is quite interesting, but way beyond the capabilities of most respondents, and of the few persons who might be ABLE to solve it, probably none of them would make the effort to solve it here, whereas they might do so if they thought their names would appear in a magazine that's actually published on paper and filed in libraries.
