For which bases b is it possible to make a pandigital number in base b that is divisible by 11 in base b?
For example, in base 4: 1023 / 11 = 33. (In base 10 this is 75 / 5 = 15.)
The divisibility rule for 11 works in other bases as well.
Consider the "odd" positioned digits as red and the "even" positioned digits as blue. The sum of reds minus the sum of blues (call it "diff") must be 0 or a multiple of 11 in base b, or (b+1).
If you start out with a test number of 01234.... ending in (b1) and compute the difference between reds and blues, "diff" is something like:
+ b/2 for even bases, and
 (b1)/2 for odd bases.
(Sorry about the leading zero, but it's easy enough to flip numbers later to remove any leading zero)
Now imagine flipping a red and blue number in their positions. The change to "diff" is always an even number, whatever amount the blue team lost, the red team gained. Thus you can only change "diff" by a + or  even number.
This explains why base 7 has no solution:
for 1023456, the smallest base 7 pandigital, diff = 3. But diff needs to be 0 or 8 or 8 and there is no way to get there in even increments starting at 3.
I used this technique to find a base 13 pandigital which is divisible by 11 in base 13.
1423056789ABC (base 13) = 30774602733042 (base 10)
which is divisible by 14 (or 11 in base 13): 2,198,185,909,503
It appears that bases which are 3 mod 4 will have no pandigital divisible by 11 in that base, or b=3+4n (n=0,1,2,...) because "diff" will be odd and no number flips can produce (b+1) which is even.
Edited on July 23, 2017, 4:56 pm

Posted by Larry
on 20170723 12:46:51 