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 thought maybe I'd tackle this case. 10 is possible as shown in my other post only if k=1 and b=17 (a=28)
So we need half (5) of the digits to sum to 17 (the other half will sum to 28.)
Here are the ways:
01259
01268
01349
01358
01367
01457
02348
02357
02456
12347
12356
Note there are eleven ways total. In each case the five digits can occupy either all the even or all the odd places. Half the time we don't have to worry about the 0 being in the front and the other half we do.
11*(5!*5!+4*4!*5!) = 285120 is the number of base 10 pandigital multiples of 11.
There are 9*9! =3265920 total pandigitals for a probability of 11/126.
This is a more than a little bit lower than 1/11 which is a bit of a surprise.
The above appears to be the start of a general formula.
(The only other case I looked at was base 4 and the probability came out to 6/18 = 1/3 which is way over.)

Posted by Jer
on 20111128 16:09:30 