N is a duodecimal (base 12) positive integer having precisely 50 digits such that each of its digits is equal to 1 except the 26th digit. If N is divisible by the duodecimal number 17, then find the digit in the 26th place.

I made a silly spreadsheet error before.

Consider a 50 digit number of all 1's.  Call it "M".
This equals (12^50 -1)/(12-1).
Mod(12^50,19) is 11.  If that has 1 subtracted, then the "mod" is 10.
Any number which is 10mod19 which is divided by 11 becomes 13mod19.
So: M is 13mod19 which agrees with Charlie's UBASIC result.

Say the 26th digit is "X"

Depending on whether the 26th digit is being counted from the left or from the right, then N equals:
either M + (X-1)*12^24   or   M + (X-1)*12^25.
Since M is 13mod19, the part after the "+" must be 6mod19.
But Mod(12^24,19) = 1   and  Mod(12^25,19) = 12
So (X-1) must be either  6 or 10
    (1*6 is 6mod19;    10*12=120 which is 6mod19)
So: X must be 7 if we are counting from the LEFT
or 11 which is "B" in base 12 if counting from the RIGHT

7 or B

Posted by Larry on 2010-05-29 16:47:27