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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 two consecutive positive integers be 2n – 1, 2n + 1 where n ≥ 1 ∈ Z.
Given that 2n – 1 < 10 and 2n + 1 < 10
∴ 2n < 11 and 2n < 9
∴ 2n < 9
∴ `"n"<9/2` ......(i)
Also, (2n – 1) + (2n + 1) > 11
∴ 4n > 11
∴ `"n">11/4` ......(ii)
From (i) and (ii)
`11/4<"n"<9/2`
Since, n is an integer.
∴ n = 3, 4
n = 3 gives 2n – 1 = 5, 2n + 1 = 7 and
n = 4 gives 2n – 1 = 7, 2n + 1 = 9
∴ The pairs of positive consecutive integers are (5, 7) and (7, 9).
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