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Divisibility question.
Hi everybody! I am new to this forum and have just joined recently. The forum seems great and I've always wanted to find and join an online finance forum Anyways, I am currently an undergraduate student in Pure Mathematics, Finance and Actuarial studies, so here's a little brainteaser combinatorics question Do there exist at least 10,000 10 digit numbers divisible by 7, all of which can be obtained from one another by a reordering of the digits?
Last edited by TrueTears on July 18th, 2011, 10:00 pm, edited 1 time in total.
Divisibility question.
The answer (counting) is yes, no clever answer (divisibility) yet, but it probably involves the number of solutions of the equation resulting from the repeated application of the divisibility rule to a 10-digit number, where the variables are the digits.
Divisibility question.
Correct, can you think of a purely combinatorial approach? That would be quite slick ;P
Divisibility question.
For the permutations of {0,1,2,...,9} with exactly k elements, there seems to be always a particular choice that produces (k-1)! numbers that are divisible by 7 (respecting no zero on the left):k=3: {1,4,7}k=4: (1,3,5,8}k=5: {1,2,7,8,9}It will take a while to check this for k>5, but it looks interesting.
Divisibility question.
The number of 10-digit numbers modulo reordering of digits (i.e., the number of orbits of the action of the symmetric group S_10 by permutation of digits) is equal to the number of monomials of degree 10 in 10 variables, i.e. C(19,10) = 92378.Therefore, by the pigeon hole principle, there must be an orbit containing at least10^10 / ( 7 * 93278 ) > 15464 numbers divisible by 7.
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- DevonFangs
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Divisibility question.
QuoteOriginally posted by: cm27874The number of 10-digit numbers modulo reordering of digits (i.e., the number of orbits of the action of the symmetric group S_10 by permutation of digits) is equal to the number of monomials of degree 10 in 10 variables, i.e. C(19,10) = 92378.Therefore, by the pigeon hole principle, there must be an orbit containing at least10^10 / ( 7 * 93278 ) > 15464 numbers divisible by 7.WTF, dude.
Divisibility question.
Consider all the integers generated by reordering the 9 digit number 123456789. Let's denote this set by S. Clearly, |S| = 9!.So, there must exist k in {0, 1, 2, 3, 4, 5, 6} such that S_k = { x in S such that x mod 7 = k } has more than 9!/7 elements. In other words, |S_k| = 9!/7 > 50000.All the 9!/7 numbers in the set 10*S_k mod 7 = 10*k. All numbers in 10*S_k has 10 digits.Find the number a in {0, 1, 2, 3, 4, 5, 6} such that a + 10*k mod 7 = 0.Now, all numbers in 10*S_k + a are divisible by 7 and are of 10 digits and are generated by reordering 123456789a.