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Find all pair of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11 .
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Verified Answer
Let $x$ be the smaller of the two odd positive integers so that the other is $x+2$. Then we should have $x < 10$
$\begin{aligned}
&\therefore \quad x+(x+2)>11 \Rightarrow 2 x+2>11 \\
&\Rightarrow \quad 2 x>11-2
\end{aligned}$
$\Rightarrow 2 x>9 \Rightarrow x>\frac{9}{2}$
Hence, if one number is 5 (odd number), then the other is seven, since $7 < 10$. If the smaller number is seven, then the other is nine. Hence, possible are $(5,7)$ and $(7,9)$.
$\begin{aligned}
&\therefore \quad x+(x+2)>11 \Rightarrow 2 x+2>11 \\
&\Rightarrow \quad 2 x>11-2
\end{aligned}$
$\Rightarrow 2 x>9 \Rightarrow x>\frac{9}{2}$
Hence, if one number is 5 (odd number), then the other is seven, since $7 < 10$. If the smaller number is seven, then the other is nine. Hence, possible are $(5,7)$ and $(7,9)$.
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