Question

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

Answer 1

Let ‘a’ be any positive even integer and ‘b= 6’.

Therefore, a = 6q +r, where 0 ≤ r < 6.

Now, by placing r = 0, we get, a = 6q + 0 = 6q

By placing r = 1, we get, a = 6q +1

By placing, r = 2, we get, a = 6q + 2

By placing, r = 3, we get, a = 6q + 3

By placing, r = 4, we get, a = 6q + 4

By placing, r = 5, we get, a = 6q +5

Thus, a = 6q or, 6q +1 or, 6q + 2 or, 6q + 3 or, 6q + 4 or, 6q +5.

But here, 6q +1, 6q + 3, 6q +5 are the odd integers.

Therefore, 6q or, 6q + 2 or, 6q + 4 are the forms of any positive even integers

Answer 2

Answer:

CAN BE PROVED

Step-by-step explanation:

6q , 6q+1

a=bq+r,    0 ≤ r less than b

∴ r=1,2,3,4,5

r=0

a=6q+0

a=6q

∴IT IS EVEN

r=1

a=6q+1

∴IT IS ODD

r=2

a=6q+2

∴IT IS EVEN(2 is remainder)

r=3

a=6q+3

∴IT IS ODD

r=4

a=6q+4

∴IT IS EVEN(4 is remainder)

r=5

a=6q+5

∴IT IS ODD

∴ANY POSITIVE INTEGER IS OF THE FORM 6q+1, 6q+3, 6q+5

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