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

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Answer 2

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We Consider An positive integer as a.

On dividing a by b .

Here ,

let q is the quotient and r is the remainder.

Now,

a = bq + r , 0 ≤ r < b ..... ( 1 )

[ by using Euclid's division lemma]

Here we putting b = 6 in eq ( 1 )

Here we find ,

a = 6q + r , 0 ≤ r < b ..... ( 2 )


so here possible values of r = 1 , 2, 3, 4, 5.

If r = 0, then find Equation (2) , a = 6q.


Here,

6q is is divisible by 2 , so 6q is here Even .

If r = 1 , then find Equation (2) , a = 6q + 1.

Here,

6q + 1 is not divisible by 2 , so 6q + 1 is here odd.

If r = 2 , then find Equation (2) , a = 6q + 2.

Here,

6q + 2 is not divisible by 2 , so 6q + 2 is here even.

If r = 3 , then find Equation (2) , a = 6q + 3.

Here,

6q + 3 is not divisible by 2 , so 6q + 3 is here odd.


If r = 4 , then find Equation (2) , a = 6q + 4.

Here,

6q + 4 is not divisible by 2 , so 6q + 4 is here even.

If r = 5 , then find Equation (2) , a = 6q + 5.

Here,

6q + 5 is not divisible by 2 , so 6q + 5 is here odd.

so , a is odd , so a cannot be 6q,6q+2, 6q+4.


Therefore any positive odd integer is of the form 6q+1 , 6q +3, 6q + 5.


Here we proved.


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