Question

Find all pairs of consecutive even positive integers, both of which are larger than 5, such that their sum is less than 23.

Answer 1

Suppose ‘x’ be the smaller of the two consecutive even positive integers. Now the other even integer is x + 2.

Given as

Here, both the even integers are greater than 5 and their sum is less than 23.

Therefore,

x > 5 and x + x + 2 < 23

x > 5 and 2x < 21

x > 5 and x < 21/2

5 < x < 21/2

5 < x < 10.5

Now, from this inequality, we can say that x lies between 5 and 10.5.

Therefore, the even positive integers lying between 5 and 10.5 are 6, 8 and 10.

Then, let us find pairs of consecutive even positive integers.

Let x = 6, then (x + 2) = (6 + 2) = 8

Let x = 8, then (x + 2) = (8 + 2) = 10

Let x = 10, then (x + 2) = (10 + 2) = 12.

x = 6, 8, 10 [Since, x is even integer]

∴ The required pairs of even positive integer are (6, 8), (8, 10) and (10, 12)