Determine the probability (in terms of n) that a basen positive integer, starting with a nonzero digit and containing each of the digits 0, 1, 2,..., n1 exactly once, is a multiple of the basen number 11.
I'm nowhere near having solved this but I can show that in some cases it is not possible. It appears to me that any formula will be extremely complicated.
The divisibility rule for 11 on base 10 is the same as in any base, n.
If the difference between the sums of the numbers in the even and odd places is a multiple of 11, the number is divisible by 11.
So we can think of the pandigital number is being split into its even and odd places and call these sums a and b (a>b).
In base n, the digits 0 through (n1) sum to n(n1)/2 = nē/2  n/2.
In base n, the number 11 is equal to n+1
so we have the system
a+b = nē/2  n/2
ab = k(n+1)
Solving for b gives b = (nēn)/4  k(n+1)/2
In different bases this may not be possible:
Base 2: b= .51.5k does not give a positive integer for nonnegative k
Base 3: b=1.5  2k never gives an integer
Base 4: b= 3  2.5k if k=0, b=3 (and a=3) which gives 6 possibilities including for example 3201
Base 5: b = 5  3k which gives two possibilities k=0, b=5 (example: 43120) and k=1, b=2 (ex: 40321)
Base 6: b = 7.5  3.5k which gives k=1, b=4 (ex: 534120)
Base 7: b = 10.5  3k never gives an integer
Base 8: b = 14  4.5k which gives k=0, b=14 (ex: 76453201) and k=2, b = 5 which is impossible since the smallest 4 digits sum to 6
Base 9: b=185k which gives k=0, b=18 (many), k=1, b=13 (many), k=2, b=8 (ex: 857251403), k=3, b=3 is too small
Base 10, b = 22.5  5.5k which gives k=1, b=17 (ex: 8071624539), k=3, b=6 which is too small.
I haven't the time to decide for which larger n it will be impossible.
Just to count the base 10 solutions you would need to find every partition of 5 numbers and a sum of 17. Then for each multiply this by 5! and by 5! excepting some that will cause a leading 0.

Posted by Jer
on 20111128 11:46:24 