Find 3 consecutive integers?

Question

Find three consecutive integers such that the product of the first and the third exceeds the product of 8 and the second by 32... 

 

Answer 1

Three consecutive integers can be represented by n, n + 1, n + 2

 

The product of the first and the third, or n(n + 2), exceeds the product of 8 and the second, 8(n + 1) by 32.

 

So, n(n + 2)  = 8(n + 1) + 32

 

We can multiply out to get

 

n^2 + 2n = 8n + 8 + 32

 

n^2 -6n - 40 = 0

 

(n - 10)(n + 4) = 0

 

So, n = 10 or n = -4

 

There are actually two sets of answers:

10, 11, 12

and

-4, -3, -2

 

Check:

(10)(12) = (8)(11) + 32

120 = 120

 

(-4)(-2) = (8)(-3) + 32

8 = 8