Math, asked by amarsinghswag16, 10 months ago

jaldi answer do yrrr vvi questions hai​

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Answered by ShahzebSheikh
0

Here is your solution,

                                                 

Let n be any arbitrary positive odd integer.

On dividing n by 6, let m be the quotient and r be the remainder. So, by Euclid’s division lemma, we have

n = 6m + r, where 0 ≤ r ˂ 6.  

As 0 ≤ r ˂ 6 and r is an integer, r can take values 0, 1, 2, 3, 4, 5.

⇒ n = 6m or n = 6m + 1 or n = 6m + 2 or n = 6m + 3 or n = 6m + 4 or n = 6m + 5 .

                                   

But n ≠ 6m or n ≠ 6m + 2 or n ≠ 6m + 4 ( ∵ 6m, 6m + 2, 6m + 4 are multiples of 2, so an even integer whereas n is an odd integer)

⇒ n = 6m + 1 or n = 6m + 3 or n = 6m + 5  Thus, any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5), where m is some integer.  

                                       

HOPE IT HELPS..............................                                

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