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Question

(A) Let $x$ be the smaller of the two consecutive odd positive integers. Then,the other integer is $x+2$.Since both integers are smaller than $10$,we

Vedclass · Accepted Answer

$(5, 7), (7, 9)$

Vedclass · Answer

(A) Let $x$ be the smaller of the two consecutive odd positive integers. Then,the other integer is $x+2$.Since both integers are smaller than $10$,we have:$x+2 $\Rightarrow x Also,the sum of the two integers is more than $11$:$x + (x+2) > 11$$\Rightarrow 2x + 2 > 11$$\Rightarrow 2x > 9$$\Rightarrow x > 4.5$ ... $(ii)$From $(i)$ and $(ii)$,we have $4.5 Since $x$ must be an odd positive integer,the possible values for $x$ are $5$ and $7$.If $x = 5$,the pair is $(5, 7)$.If $x = 7$,the pair is $(7, 9)$.Thus,the required pairs are $(5, 7)$ and $(7, 9)$.

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

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