What is value of $a+b+c+d+e$?

Question

What is value of $a+b+c+d+e$? If given :

$$abcde=45$$

And $a,b, c, d, e$ all are distinct integer.

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My attempt :

I calculated, $45 = 3^2 \times 5$.

Can you explain, how do I find the distinct values of $a,b, c, d, e$ ?

Answer 1

So $$45=5\cdot(\pm 3)^2\cdot(\pm1)^2$$ which implies that the only possibility (up to renaming) is $a=1, b=-1, c=3, d=-3, e=5$.

Answer 2

The divisors of $45$ are:

$$\pm 1 \qquad \pm 3 \qquad \pm 5 \qquad \pm 9 \qquad \pm 15 \qquad \pm 45 $$

The only way to multiply $5$ distict ones and make $45$ is

$$(-1)\cdot 1\cdot(-3)\cdot 3\cdot 5$$ So the sum is $5$.

Answer 3

Indeed... the values would be $5,3,-3,-1,1$. So adding them should give you $5$.