How many sets of three consecutive integers are there in which the sum of the three integers equals their product?

3

Step-by-step explanation:

since the 3 integers are consecutive, we are dealing with x, x+1, x+2.

and their sum is the same as their product :

x + (x + 1) + (x + 2) = x(x + 1)(x + 2)

3x + 3 = x(x² + 3x + 2) = x³ + 3x² + 2x

x³ + 3x² - x - 3 = 0

this is a polynomial of third degree.

and as such it has 3 solutions.

of course, it could be that some of them are the same or are even in the realm of complex numbers (i = sqrt(-1)), but usually these 3 solutions are different real numbers.

I tried x=1 just to see, and, hey, it is a solution for this equation.

x = 1 means that the other 2 consecutive integers are 2 and 3.

and indeed, 1+2+3 = 1×2×3 = 6.

now it is easier to find the other 2 solutions, as a zero solution can be expressed as a factor of the whole expression.

for x = 1 the factor term is (x - 1), as this term is then turning 0, when x = 1.

I can divide the main expression by this factor and then analyze the quotient about the other 2 solutions.

x³ + 3x² - x - 3 : x - 1 = x² + 4x + 3

- x³ - x²

0 4x² - x

- 4x² - 4x

0 + 3x - 3

- 3x - 3

0 0

so, the original expression can be written as

(x² + 4x + 3)(x - 1).

now we need to find the 2 zero solutions for x²+4x+3

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 4

c = 3

so,

x = (-4 ± sqrt(4² - 4×1×3))/(2×1) =

= (-4 ± sqrt(16 - 12))/2 = (-4 ± sqrt(4))/2 =

= (-4 ± 2)/2 = -2 ± 1

x1 = -2 + 1 = -1

x2 = -2 - 1 = -3

so, we have the additional solutions :

-1 0 1

-3 -2 -1

-1 + 0 + 1 = -1×0×1 = 0

-3 + -2 + -1 = -3×-2×-1 = -6

and there we have it fully proven :

there are 3 different sets of 3 consecutive integers with the same sum as product.

the answer is d on edge

step-by-step explanation: