Question

Find three consecutive positive integers such that sum of square of first and product of other two is 277.

Answer 1

Let the three consecutive positive integers be:

x, x + 1, x + 2

Square of the first number: x²

Product of the other two numbers: (x + 1) (x + 2)

Given:

x² + (x + 1) (x + 2) = 277

x² + (x² + 3x + 2) = 277

2x² + 3x + 2 = 277

2x2 + 3x − 275 = 0

Factor:

(2x + 25) (x − 11) = 0

`x = -25/2 or x = 11`

Since the numbers must be positive integers, take x = 11.

The three consecutive positive integers are: 

11, 12, 13​