Question

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

Answer 1

Let $x$ be the smaller of the two consecutive odd positive integers.

Then, the other integer will be $x+2.$
Since both the integers are smaller than $10$,
$\Rightarrow x+2<10$
$\Rightarrow x<10-2$
$\Rightarrow x<\mathrm{8...}\left(i\right)$
Also, the sum of the two integers is more than $11$.
$\therefore x+(x+2)>11$
$\Rightarrow 2x+2>11$
$\Rightarrow 2x>11-2$
$\Rightarrow 2x>9$
$\Rightarrow x>\frac{9}{2}$
$\Rightarrow x>4.5$ ...(ii)
From (i) and (ii) since $x$ is an odd number ,$x$ can take the values $5$ and $7$. Thus the required possible pairs are $(5,7)$ and $(7,9)$