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 50 - Digit Number II (Posted on 2010-05-29)
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.

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes)

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 Not sure if this is correct. | Comment 3 of 9 |

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

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