Re: [math-fun] Pandigital Trivia Spoiler
Hans Havermann <gladhobo@bell.net> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
What do 3076521984, 3718250496, and 6398410752, and no other numbers, have in common?"
They are obviously pandigital numbers and thus share the divisor 3^2. The GCD of the three numbers divided by 9 is a large power of 2. A quick check shows that no other pandigital has that large a power-of-two divisor (the next largest being for 9805234176).
Correct. My three numbers are all divisible by 2^21. No 10-digit base-10 pandigital numbers are divisible by 2^22. I've looked into the highest powers of small primes among the n-digit base-n pandigital numbers for small n. Here's a table. As always with tables, it should be viewed in a fixed-width font. base: 2 3 4 5 6 7 8 9 10 11 12 13 2: 1 0 3 1 11 0 5 2 21 0 29 1 3: 0 1 3 5 7 9 10 3 15 17 18 21 5: 0 1 2 1 5 5 6 9 9 11 12 13 7: 0 1 2 2 3 1 7 6 8 9 9 13 11: 0 1 1 2 2 3 5 7 5 1 8 9 13: 0 0 1 2 2 3 3 5 5 6 8 1 I can easily push it to higher primes, but not to higher bases. Computing the entries for base n will take roughly n times longer than computing them for all smaller bases put together. I may extend this table to powers of composite numbers. It should be obvious why the entry equals 1 whenever the row and column numbers are equal. The "opposite" problem, finding n-digit base-n pandigital primes, isn't very interesting. There are only three of them. Can you find them? (I haven't looked into non-standard bases, such as negabinary, Fibonacci, or balanced ternary.) Which of these rows and columns would be of interest to OEIS? None of them appear to be there yet.
I find the pattern in the first row of your table very interesting. Is there any intuitive reason that pandigital numbers in odd bases only have very small powers of two as divisors? On Tue, Mar 31, 2020 at 7:56 PM Keith F. Lynch <kfl@keithlynch.net> wrote:
Hans Havermann <gladhobo@bell.net> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
What do 3076521984, 3718250496, and 6398410752, and no other numbers, have in common?"
They are obviously pandigital numbers and thus share the divisor 3^2. The GCD of the three numbers divided by 9 is a large power of 2. A quick check shows that no other pandigital has that large a power-of-two divisor (the next largest being for 9805234176).
Correct. My three numbers are all divisible by 2^21. No 10-digit base-10 pandigital numbers are divisible by 2^22.
I've looked into the highest powers of small primes among the n-digit base-n pandigital numbers for small n. Here's a table. As always with tables, it should be viewed in a fixed-width font.
base: 2 3 4 5 6 7 8 9 10 11 12 13
2: 1 0 3 1 11 0 5 2 21 0 29 1 3: 0 1 3 5 7 9 10 3 15 17 18 21 5: 0 1 2 1 5 5 6 9 9 11 12 13 7: 0 1 2 2 3 1 7 6 8 9 9 13 11: 0 1 1 2 2 3 5 7 5 1 8 9 13: 0 0 1 2 2 3 3 5 5 6 8 1
I can easily push it to higher primes, but not to higher bases. Computing the entries for base n will take roughly n times longer than computing them for all smaller bases put together.
I may extend this table to powers of composite numbers.
It should be obvious why the entry equals 1 whenever the row and column numbers are equal.
The "opposite" problem, finding n-digit base-n pandigital primes, isn't very interesting. There are only three of them. Can you find them? (I haven't looked into non-standard bases, such as negabinary, Fibonacci, or balanced ternary.)
Which of these rows and columns would be of interest to OEIS? None of them appear to be there yet.
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-- Andy.Latto@pobox.com
I'm making a very big free-associative leap, but this topic seems vaguely related to something I've thought of as "reverse long division". I'd like to know whether it's been studied; I think the problem under discussion is in the same conceptual area. If we have a big number N written out in a radix r, and we want to divide it by some much smaller d, the familiar long-division algorithm is the obvious tool. But if we know in advance that N is an exact multiple of d, then there is another algorithm that works -- perhaps only for certain values of d. Suppose we want to divide 9044 by 7, and we have been told in advance that 7 is in fact a divisor of 9044, that is, that 7Q = 9044. Then we know that Q's last digit must be 2; no other final digit could produce a product ending in 4. Subtracting 14 from 9044, we obtain 9030, and the rest of the problem is to divide 903 by 7. The terminal 3 is only possible if the tens digit of Q is 9. Subtract 63 from 903, we get 840, and now we must divide 84 by 7. The final 4 forces the hundreds digit of Q to be 2, and in the last step we are left with 84-14=70, and must divide 7 by 7; so the answer is reconstructed from the digits as Q = 1292. I believe exact multiples of 7 can always be divided by 7 from right to left in this fashion. There is a connection to the theory of r-adic numbers; I'm using r instead of p here because the radix does not need to be prime, but I think r and d need to be relatively prime for this procedure to work. Now suppose we are to find a pandigital number in an odd base, which we are told is a multiple of a particular given power of 2. A very similar procedure allows us to put very strong constraints on the digit sequence starting from the right. If we just guess the last digit, the process tightly constrains the higher-order digits. I think the answer to Andy's question might yield to this style of analysis. On Tue, Mar 31, 2020 at 11:20 PM Andy Latto <andy.latto@pobox.com> wrote:
I find the pattern in the first row of your table very interesting. Is there any intuitive reason that pandigital numbers in odd bases only have very small powers of two as divisors?
On Tue, Mar 31, 2020 at 7:56 PM Keith F. Lynch <kfl@keithlynch.net> wrote:
Hans Havermann <gladhobo@bell.net> wrote:
"Keith F. Lynch" <kfl@KeithLynch.net> wrote:
What do 3076521984, 3718250496, and 6398410752, and no other numbers, have in common?"
They are obviously pandigital numbers and thus share the divisor 3^2. The GCD of the three numbers divided by 9 is a large power of 2. A quick check shows that no other pandigital has that large a power-of-two divisor (the next largest being for 9805234176).
Correct. My three numbers are all divisible by 2^21. No 10-digit base-10 pandigital numbers are divisible by 2^22.
I've looked into the highest powers of small primes among the n-digit base-n pandigital numbers for small n. Here's a table. As always with tables, it should be viewed in a fixed-width font.
base: 2 3 4 5 6 7 8 9 10 11 12 13
2: 1 0 3 1 11 0 5 2 21 0 29 1 3: 0 1 3 5 7 9 10 3 15 17 18 21 5: 0 1 2 1 5 5 6 9 9 11 12 13 7: 0 1 2 2 3 1 7 6 8 9 9 13 11: 0 1 1 2 2 3 5 7 5 1 8 9 13: 0 0 1 2 2 3 3 5 5 6 8 1
I can easily push it to higher primes, but not to higher bases. Computing the entries for base n will take roughly n times longer than computing them for all smaller bases put together.
I may extend this table to powers of composite numbers.
It should be obvious why the entry equals 1 whenever the row and column numbers are equal.
The "opposite" problem, finding n-digit base-n pandigital primes, isn't very interesting. There are only three of them. Can you find them? (I haven't looked into non-standard bases, such as negabinary, Fibonacci, or balanced ternary.)
Which of these rows and columns would be of interest to OEIS? None of them appear to be there yet.
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-- Andy.Latto@pobox.com
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participants (3)
-
Allan Wechsler -
Andy Latto -
Keith F. Lynch