Given that x is a positive integer less than 6, how many values can k take on such that 3x = k (mod9) has no solutions in x?
If 3x=k (mod 9) has a solution, then 3x - k = 9m, for some integer m. That is k = 3x - 9m = 3(x-3m), which implies that k is a multiple of 3. But k is the remainder upon division of any of the numbers in the set {3, 6, 9, 12, 15} by 9, so k =0, 3, 6 are the only values k can take, if the given equation has a solution. But in general division of a integer by nine can result in 0, 1, 2, 3, ..., 8 left over as remainders; so 1, 2, 4, 5, 7, 8 are the values k can take if the equation has no solutions.
