Question

732 Show that any positive odd integer 6m + 5 is of the form 3n + 2 where m in Z,n in Z

Answer 1

Answer:

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 m is some integer.

Answer 2

Explanation:

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