shortcut technique to solve algebric problem

Question

If we multiply three consecutive numbers, we get 120. what the summation of those numbers? 

Is there any shortcut way of solving this problem without doing much calculation?

Answer 1

Factoring, $120=2^3\cdot3\cdot5$.

We need consecutive factors. At a glance, they could be $3,4,5$ or $4,5,6$.

Hope this fits in your requirements.

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With the help of a calculator:

You can observe that $(n-1)n(n+1)=n^3-n$ is close to $n^3$. As $\sqrt[3]{120}=4.93\cdots$, you can conjecture $n=5$.

Answer 2

Three consecutive integers can be written as $x-1, x, x+1$. Thus, we have: $$(x-1)x(x+1)=120$$ First, note that $x > 0$ because otherwise, the product would be $0$ or negative since the product of three negative consecutive integers would be negative. $$x^3-x-120=0$$ Use synthetic division with positive integer roots in order to find the answer.

Answer 3

Let $n$ be the middle of the three consecutive numbers.

They tell us that $(n-1)\cdot n\cdot (n+1) = 120$

We want to find $(n-1)+n+(n+1)$

This is more convenient in my opinion than letting the numbers be $n,n+1,n+2$ since things simplify nicely.

The statements simplified are:

if $n^3-n=120$, then find $3n$

As for continuing to solve, one could use heavier tools like cardano's method, or you could make the assumption that since the question was being asked that you expect there to be a positive integer value $n$ such that this is true. Notice that $3^3=27, 4^3=64, 6^3=216$, so $4^3-4$ is too small to be $120$ and $6^3-6$ is too big to be $120$... This implies that...