Question

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

Answer 1

Hii friend,

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.

HOPE IT WILL HELP YOU..... :-)

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let n be the given positive odd integer.

On dividing n be 6,

Assume m be the quotient

And r be the remainder.

Then,

By Euclid's division lemma, we have

n = 6m + r, where 0 ≤ r < 6

⇒ n = 6m + r,

where r = 0, 1, 2, 3, 4, 5

⇒ 6m or (6m + 1) or (6m + 2) or (6m + 3) or (6m + 4) and (6m + 5).

But, n = 6m, (6m + 2), (6m + 4) give even values of n.

Hence, any positive odd integer is of the form (6m + 1) or (6m + 3) or (6m + 5) Where n is some integer.