Binomial Theorem

Question
CBSEENMA11014373

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

Solution

Let the two consecutive odd positive integers be x and x + 2
Since both the integers are smaller than 10
∴                        x < 10 and x + 2 < 10
rightwards double arrow                     x < 10   and    x < 8
rightwards double arrow                      x < 8                                                                                              ...(i)
Also,  straight x plus left parenthesis straight x plus 2 right parenthesis greater than 11 space rightwards double arrow space 2 straight x plus 2 space greater than 11 space rightwards double arrow space 2 straight x greater than 9 space rightwards double arrow space straight x greater than 9 over 2                                     ...(ii)
From (i) and (ii), we have
             9 over 2 less than straight x less than 8 comma                   space space space straight x space element of space straight Z space rightwards double arrow space straight x space equals space 5 comma space 6 comma space 7
But x is odd.
rightwards double arrow x = 5, 7 and x + 2 = 7, 9
Hence, the pair of odd numbers are satisfying the given conditions are 5, 7 and 7, 9.

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