Question

show that any positive odd integer is of the form (6 m + 1 )or (6 m + 3) or (6 m + 5) .Where M is some integer

Answer 1

Answer:

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Let n be a given positive odd integer.

On dividing n by 6 , let m be the Quotient and r be the remainder.

Then, by Euclid division lemma, we have

Dividend = Divisor × Quotient + Remainder

n = 6m + r. where r = 0 , 1 ,2 , 3 ,4 ,5

n= 6m + 0 = 6m. [ r =0]

n = 6m + 1 = 6m+1 [ r = 1 ]

n = 6m +2 = 6m +2 [ r = 2 ]

n = 6m +3 = 6m+3 [ r = 3 ]

n = 6m +4 = 6m+4 [ r = 4 ]

n = 6m+5 = 6m+5 [ r = 5 ]

N = 6m , (6m +2) , (6m+4) is even value of n.

Therefore,

when n is odd , it is in the form of (6m+1) , (6m+3) , (6m+5) for some integer m.

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Answer 2

Answer:

here is your answer

hope it will help you. ...