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 the smaller odd positive integer be x. Then, the other odd positive integer shall be x + 2.
Therefore, as per the given conditions:
$x+2<10$ and $x+x+2>11$
$\Rightarrow x<8$ and $2 x>9$
$\Rightarrow x<8$ and $x>\frac{9}{2}$
Since $x$ is an odd integer,
Therefore, $x=5,7$
Hence, pairs are $(5,7),(7,9)$.
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