P(N) denotes the product of the digits of a 2digit duodecimal (base 12) positive integer N consisting entirely of nonzero digits.
Find all possible values of a positive integer constant C > 1 such that:
N = C*P(N)
I realized this could be solved quickly with a program so I wrote one.
Let the digits be X and Y
we require 12X+Y=C*X*Y
or C=(12X+Y)/(X*Y)
where X and Y are each from 1 to 11
solutions in duodecimal:
11 = 11*1*1
12 = 7*1*2
13 = 5*1*3
14 = 4*1*4
16 = 3*1*6
28 = 2*2*8

Posted by Jer
on 20160612 17:51:42 