Question

using euclid's lemma, prove that any odd +ve integer is of the form 6m+1,6m+3,6m+5

Answer 1

Let n be a odd integer.
On dividing n by 6 , Let m be the quote and r be the remainder.
Then,
by Euclid's division lemma we have,
n = 6m+r where r =0, 1,2,3,4,5
n = 6m
n = 6m+1 , 6m+2, 6m+3,6m+4, 6m+5.
N = 6m, 6m+2,6m+4 is even values of n.
Thus,
When n is odd , it is in the form of 6m+1, 6m+3 or 6m+5 for some integer M.

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Answer 2

n = 6m+r , where 0<_r<6 .


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

Where r = 0

n = 6m --------even .

where r = 1

n= 6m+1 ------------odd.

Where r = 2

n = 6m+ 2 ---------even


Where r = 3

n= 6m+ 3------------odd.


Where r = 4

n= 6m+ 4-'----------even .

Where r = 5

n= 6m+5 ----------odd.

But , n = 6m , 6m+1, 6m+4 Are even values of n .


Thus when n is odd it form 6m+1 or 6m+2 or 6m+3 for some integer n .