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

We need to assume two consecutive even positive integers.

So, let the smaller even positive integer be x.

Then, the other even positive integer will be (x + 2).

Given: Both these numbers are larger than 5. …(i)

And their sum is less than 23. …(ii)

So,

From given statement (i),

x > 5 …(iii)

x + 2 > 5

x > 5 – 2

x > 3

Since, the number must be larger than 5, x > 3 can be ignored.

From given statement (ii),

Sum of these two consecutive even positive integers < 23

(x) + (x + 2) < 23

x + x + 2 < 23

2x + 2 < 23

2(x + 1) < 23

x + 1 < 11.5

x < 11.5 – 1

x < 10.5 …(iv)

From inequalities (iii) & (iv), we have

x > 5 & x < 10.5

It can be merged and written as

5 < x < 10.5

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

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

Now, 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.

Hence, all such pairs of consecutive even positive integers required are (6, 8), (8, 10) and (10, 12).

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