At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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

Sagot :

Answer:

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.