Form three arithmetic sequences:
1.- Numbers that are divisible by $\,4\,$ and less or equal than $\,400\,$:
$$4,8,12,...\Longrightarrow a_1=4\,\,,\,d=4\Longrightarrow a_n=a_1+(n-1)d=4+(n-1)4\leq 400\Longrightarrow n\leq 100$$
2.- Numbers that are divisible by $\,6\,$ and less or equal than $\,400\,$:
$$6,12,18,...\Longrightarrow a_1=6\,\,,\,d=6\Longrightarrow a_n=6+(n-1)6\leq 400\Longrightarrow n\leq66$$
3.- Numbers that are divisible by $\,4\,$ and $\,6\,$ and less or equal than $\,400\,$:
$$12,24,36,...\Longrightarrow a_1=12\,\,,\,d=12\Longrightarrow a_n=12+(n-1)12\leq 400\Longrightarrow n\leq 33$$
Well, take it now from here: how many positive integers less than $\,400\,$ are not divisible by $4\,,\,6\,$?
Let such a number be $10x+y$. It is divisible by $x$ and $y$.
So $(10x+y)/x$ = $ 10+ y/x$ = $10 + m $ should be a natural number where $m=y/x$.
Similarly $(10x+y)/y$ = $ 10(x/y) + 1$ = $10/m + 1 $ should be a natural number.
This is only possible for $m = 1, 2$ and $5$. So the possible numbers are
For $m=1$: No.s = $\{11,22,33,...99\}$
For $m=2$: No.s = $\{12,24,36,48\}$
For $m=5$: No.s = $\{15\}$.
Hence there are 14 such numbers
Best Answer
Obviously the equation $(9-a)+(9-b)+(9-c)=6$ is equivalent to $a+b+c=21$, but there's a trick here. Now substitute $\alpha=9-a$, $\beta=9-b$, $\gamma=9-c$. Then the equation becomes $\alpha+\beta+\gamma=6$, still with the constraint that they are digits (ie numbers between $0$ and $9$ inclusive).
The trick here is that the later equation you don't have to bother with the constraint that they are $\le 9$ since that's implied by the equation. So the only constraint left to bother with is that they are integers $\ge 0$. This means that it's supposed to be easier to see that it's a triangle number $1+2+3+4+5+6+7=28$.
Otherwise of course the direct approach is doable too, but then you have to take into account that they must be $\le 9$ too. You see that this means that $0\le b+c\le 18$ so we must have $a = 21-b-c \ge 21-18 = 3$ (the other side doesn't matter as we have $a\le 9$). Then depending on $a$ you count the number of solutions to $b+c = 21-a$ and you'll get the same sum.
Of course if you have a lot of time or a computer you can simply list the numbers and count them.