Is my method correct to answering this question? Is there a quicker method to solve this question? 
What is the smallest number that can be written as the sum of three, four and five consecutive numbers?

I encountered this question while doing my Math summer homework. I have tried to make progress on this question.
Sum of three consecutive numbers = $x + x+1 + x+2 = 3x+3$
Sum of four consecutive integers = $x + x+1 + x+2 + x+3 = 4x+6$
Sum of five consecutive integers = $x + x+1 + x+2 + x+3 + x+4 = 5x+10$
The number must be the lowest common multiple of $3x+3$ , $4x+6$ and $5x+10$, which is $60x + 30$.
Substituting $x = 0$ gives us the lowest positive, non-zero and whole number, which is $30$.
$$30 = 9 + 10 + 11\\
30 = 6 + 7 + 8 + 9\\
30 = 4 + 5 + 6 + 7 + 8$$
Is my answer correct? If not, where in my method have I produced an error? Is there an ever quicker method to solve this question?
 A: Your final answer is fine, but your method looks highly suspect at one particular line which needs clarification.  You write:

The number must be the lowest common multiple of $3x+3$ , $4x+6$ and
$5x+10$, which is $60x + 30$.

But it’s not clear what you mean by this, as the lowest common multiple of those three numbers is not $60x+30$ (try it for $x=2$).  Maybe what you’re actually doing to arrive at $60x+30$ is correct, or maybe it isn’t — but at a minimum you need to explain what you’re doing because “lowest common multiple” doesn’t describe it adequately, and the reuse of the same $x$ for three different values is a big red flag.
My hunch is that you’re taking the separate LCMs of the coefficients $(3,4,5)$ and of $(3,6,10)$ and then gluing them back together.  If so then this that’s definitely wrong.  Consider the slight variation where 3,4,5 are replaced by 2,3,5.  Then you’d have $2x+1, 3x+3, 5x+10$ combining to $30x+30$, which never works (it’s even, so it can never be the sum of two consecutive integers).  Being a multiple of $2x+1$ is very different from being. equal to $2x+1$.
This broken method lucks out sometimes because of the fact that it basically works for odd coefficients (when $k$ is odd, being the sum of $k$ consecutive integers is the same as being divisible by $k$).  But if that’s really what you’re doing then it is broken and should not be part of your solution.
A: If we are talking about a non-negative integer, I have another method to propose.


Sum of three consecutive numbers = $++1++2=3+3$


Sum of four consecutive integers = $++1++2++3=4+6$


Sum of five consecutive integers = $++1++2++3++4=5+10$

Another way of formulating a solution from this step can be as follows:
$y = 3x_1 + 3 = 4x_2+6 = 5x_3+10$
$y= 3x_1+3\equiv 0 \pmod{3}$
$y= 4x_2+6\equiv 2 \pmod{4}$
$y= 5x_3+10\equiv 0 \pmod{5}$
$y$ is divisible by $5$ and $3$ which gives us $y=15k$
The remainder left from dividing $y$ by $4$ is $2$. $y=15$ doesn't satisfy this condition, but $y=30$ does.
A: Your reasoning contains a subtle (essentially notational) error, which does not greatly affect the result you obtain. You say:

The number must be the lowest common multiple of $3x+3$, $4x+6$ and $5x+10$, which is $60x + 30$.

But the value of $x$ in each of $3x+3$, $4x+6$ and $5x+10$ is not the same; by your own results it is either $4$ or $6$ or $9$. So saying $60x+30$ and $x=0$ is not logically consistent: you did not mean to imply that the sum $3x+3=3$ was equal to $4x+6=6$ or $5x+10=10$.
The better reasoning goes as follows: One of the sums has $3$ as a factor, another has $2$ (but not $4$) as a factor, and the third has $5$ as a factor. Since they each equal the same sum, that sum must have $2,3,5$ as factors, and so must be an odd multiple of $30$. In your terms, $S=60k+30$ where $k$ is not identically the same as the $x$ in any of your sums.
$30$ itself is in fact a solution, and so are $90,150,\dots$, viz:
$$90=29+30+31=21+22+23+24=16+17+18+19+20 \\
150=49+50+51=36+37+38+39=28+29+30+31+32$$
The smallest such number is $30$, as you found.
As I get set to post this, I see that Erick Wong has made essentially the same objection.
