Question

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

Answer 1

let a be any positive interger.
On dividing a by 6,let q be the quotient and r be the remainder.

Then by Euclid's algorithm,we have:
a=6q+r,where 0≤r<6
∴a=6q+r,where r=0,1,2,3,4or5

1st case: when r=0
a=6q+0
a=6q
a=2(3q)
a=2m [where m=3q]
which is clearly an even no.

2nd case:when r=1
a=6q+2
a=2(3q)+1
a=2m+1 [where m=3q]
which clearly an odd no.

3rd case:when r=2
a=6q+2
a=2(3q+1)
a=2m [where m=3q+1]

which is an even number

4th case: when r=3
a=6q+3
a=6q+2+1
a=2(3q+1)+1
a=2m+1 [where m=3q+1]

which cleraly an odd number


5th case:when r=4
a=6q+4
a=2(3q+2)
a=2m [where m=3q+2]

which is an even number


6th case:when r=5
a=6q+5
a=6q+4+1
a=2(3q+2)+1
a=2m+1 [where m=3q+2]

which is an odd number


Also, any posotive integer can of form,
6q,6q+1,6q+2,6q+3,6q+4,6q+5,


Thus, any positive integer is of the form 6q+1,6q+3 and 6q+5

Answer 2

Euclid’s division algorithm, for two positive  integers a and b, we have

a = bq + r, 0 ≤ r < b

Let b = 6,

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

Hence

a = 6q, 6q + 1, 6q + 2, 6q + 3,

6q + 4, 6q + 5

Clearly, a = 6q, 6q + 2, 6q + 4 are even and divisible by 2.

But 6q + 1, 6q + 3, 6q + 5 are odd and not divisible by 2.

Any positive odd integer is of the form 6q + 1,  6q + 3 or 6q + 5.