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.
If my modulo math is right, and I'm not certain it is, then there is no solution.
Consider a 50 digit number of all 1's. Call it "M".
This equals (12^50 -1)/(12-1).
Mod(12^50,19) is 15. If that has 1 subtracted, then the "mod" is 14.
Any number which is 14mod19 which is divided by 11 becomes 3mod19.
So (if I'm right so far): M is 3mod19.
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 3mod19, the part after the "+" must be 16mod19.
But Mod(12^24,19) = 1 and Mod(12^25,19) = 12
So (X-1) must be either 16 or 14
(1*16 is 16mod19; 14*12=168 which is 16mod19)
So X must be 17 or 15 which does not exist in base 12
So, no such 50 digit number exists.
Posted by Larry
on 2010-05-29 15:25:52