Welcome 2016! Part 28

Let $x=y+a$ , where $y$ is the integer part while $a$ is the decimal part of $x$ .

Note that for $0<a<\frac{1}{3}$ , $\lfloor x \rfloor\ + \lfloor 2x \rfloor + \lfloor 3x \rfloor = (y) + (2y) + (3y)$ , which sums to $6y$ .

For $\frac{1}{3} \leq a < \frac{1}{2}$ , $\lfloor x \rfloor\ + \lfloor 2x \rfloor + \lfloor 3x \rfloor = (y) + (2y) + (3y+1)$ , which sums to $6y+1$ .

For $\frac{1}{2} \leq a < \frac{2}{3}$ , $\lfloor x \rfloor\ + \lfloor 2x \rfloor + \lfloor 3x \rfloor = (y) + (2y+1) + (3y+1)$ , which sums to $6y+2$ .

For $\frac{2}{3} \leq a < 1$ , $\lfloor x \rfloor\ + \lfloor 2x \rfloor + \lfloor 3x \rfloor = (y) + (2y+1) + (3y+2)$ , which sums to $6y+3$ .

From above, it is apparent that every integer in the form of $6y, 6y+1, 6y+2$ and $6y+3$ can be represented as required while integers of the form $6y+4$ and $6y+5$ cannot be represented.

A quick computation reveals that there are $\boxed{67}$ such integers.