The product of the second and third of three consecutive integers is 2 more than 13 times the first integer. Find the integers.

Answer 1

Here's what we know:

a = b - 1

c = b + 1 (consecutive integers mean the three numbers are next to each other on a number line)

bc = 13a + 2 (the product of the latter two is equal to 13 times the former plus another 2)

We have values for our variables, so plug them in:

b(b + 1) = 13(b - 1) + 2

distribute:

b² + b = 13b - 13 + 2

b² + b = 13b - 11

I see a quadratic forming here. Let's move everything to one side:

b² - 12b + 11 = 0

We now need two numbers that when added together equals 11 and multiplied equals -12:

(b + 12)(b - 1) = 0

b = -12, 1

so, in turn, c could be -11 or 2, and a could be -13 or 0

Let's check the possibilities to see if either one or both are possible:

-12 * -11 = -13 * 13 + 2

132 = -167

yeah, no

1 * 2 = 13 * 0 + 2

2 = 0 + 2

2 = 2

so a = 0, b = 1, and c = 2