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 is x + 2.

Since both the integers are smaller than 10,

x + 2 < 10

⇒ x < 10 – 2

⇒ x < 8  ...(i)

Also, the sum of the two integers is more than 11.

∴ x + (x + 2) > 11

⇒ 2x + 2 > 11

⇒ 2x > 11 – 2

⇒ 2x > 9

= `x > 9/2`

= x > 4.5     ...(ii)

From (i) and (ii), we obtain.

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).

Answer 2

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).