Question

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

Answer 1

According the question,

Let the smaller odd natural number be $x$

Since the larger integers is consecutive be $x$

it will be $x$ + 2

Given that,

both integers are larger then 10,

i.e, $x$ > 10

& $x$ + 2 > 10

$x$ > 10 - 2

$x$ > 8

Since $x$ >10 & $x$ > 8

$x$ >10

Also, gib=ven that

Sum of the numbers is less then 40

$x$ + ($x$ +2) < 40

2$x$ + 2 < 40

2$x$< 40 - 2

2$x$ < 38

$x$<38/2

$x$<19

Hence,

$x$ > 10 & $x$ < 19

Natural number greater then 8 but less then 19 are 11, 12, 13, 14, 15, 16, 17, 18

Odd natural numbers are 11, 13, 15, 17

Hence possible pairs are follow,

(11, 13), (13, 15), (15, 17), (17, 19)