Question

show that any positive odd integer is of the form 6q + 1 , 6q + 3 , 6q+5 , where q is some integer.

Answer 1

Now
By Euclids division lemma we know that positive odd integer is of the form 6q+r where 0 <_ r <6
now therefore
Now we will get the following results:-
a=6q+r
1) a=6q---------->even
2) a=6q+1-------->odd
3)a=6q+2---------> even
4)a=6q+3--------->odd
5)a=6q+4---------->even
6)a=6q+5---------->odd
Thus from the above results we can say that any positive odd integer is of the form 6q+1,6q+3 or 6q+5 for some integer q
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Answer 2

HEY FRIEND HERE IS UR ANSWER,

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd Therefore, any positive odd integer is of the form of 6q+1, 6q+3 and 6q+5, where q is some integer.

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