Find all pairs of consecutive odd positive integers both of which are smaller

Question:

Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Solution:

Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2.

Since both the integers are smaller than 10,

$x+2<10$

$\Rightarrow x<10-2$

$\Rightarrow x<8 \ldots$ (i)

Also, the sum of the two integers is more than 11.

$\therefore x+(x+2)>11$

$\Rightarrow 2 x+2>11$

$\Rightarrow 2 x>11-2$

 

$\Rightarrow 2 x>9$

$\Rightarrow x>\frac{9}{2}$

 

$\Rightarrow x>4.5$    (ii)

From (i) and (ii), we obtain $4.5

Since $x$ is an odd number, $x$ can take the values, 5 and 7 .

 

Thus, the required possible pairs are $(5,7)$ and $(7,9)$.

Leave a comment