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Question

For some positive integer n, every positive odd integer is of the form

(a) n
(b) n + 1
(c) 2n
(d) 2n + 1


Solution

(d) 2n + 1

Every positive odd integer is of the form 2n + 1
Proof: 
            Let n be the given positive odd integer.
On dividing n by 2, let q be the quotient and r be the remainder.
On applying Euclid's algorithm, we get:
          n = 2q + r, where 0 ≤ r ≤ 2
    ⇒  n = 2q + r, where r = 0, 1
    ⇒  n = 2q and n = 2q + 1
But,
        n = 2q = even
Thus, when n is odd, it is in the form of n = 2q + 1.

Mathematics
RS Aggarwal (2015)
Standard X

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