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?
Best Answer
Your final answer is fine, but your method looks highly suspect at one particular line which needs clarification. You write:
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.