Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.

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Find all pairs of consecutive odd natural number, both of which are larger than 10, such that their sum is less than 40.

Solution:

Let ‘x’ be the lesser of the two odd natural numbers that follow. x + 2 is the other odd number.

According to the question, the sum of the natural numbers is less than 40, and they are both bigger than 10. So,

$x\text{ }>\text{ }10\text{ }and\text{ }x\text{ }+\text{ }x\text{ }+\text{ }2\text{ }<40$

$x\text{ }>\text{ }10\text{ }and\text{ }2x\text{ }<\text{ }38$

$x\text{ }>\text{ }10\text{ }and\text{ }x\text{ }<\text{ }38/2$

$x\text{ }>\text{ }10\text{ }and\text{ }x\text{ }<\text{ }19$

$10\text{ }<\text{ }x\text{ }<\text{ }19$

As a result of this inequality, we can deduce that x is somewhere between 10 and 19.

So, between 10 and 19, the odd natural numbers are 11, 13, 15, and 17. (With the exception of 19 as x 19)

Let’s look for pairs of odd natural numbers that are sequential.

When x = 11, then

(x + 2) = (11 + 2) = 13

When x = 13, then

(x + 2) = (13 + 2) = 15

When x = 15, then

(x + 2) = (15 + 2) = 17

When x = 17, then

(x + 2) = (17 + 2) = 19.

x = 11, 13, 15, 17 [Since, x is an odd number]

(11, 13), (13, 15), (15, 17), and (17, 19) are the required pairings of odd natural numbers.

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