Question

show that any '+ve' odd integer is of the form 6q+1 or 6q+3 or 6q+5​

Answer 1

Let a be a given integer.

On dividing a by 6 , we get q as the quotient and r as the remainder such that

a = 6q + r,

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

When r=0

$a = 6q\\= 2(3q) \ even$

When r=1

$a = 6q + 1, \ odd$

When r=2

$a = 6q + 2, \\= 2(3q+1) \ even$

When r = 3

$a=6q + 3, \ odd$

When r=4

$a=6q + 4\\= 2(3q+2) \ even$

When r=5,

$a= 6q + 5 ,\ odd$

$\boxed{\boxed{\bold{Therefore, \ Any \ positive \ odd \ integer \ is \ of \ the \ form \ 6q+1, \ 6q+3 \ or \ 6q+5.}}}}$

                                                                     

Answer 2

$</p><p>\tt{\huge{\pink{Hello!!! }}}$

Suppose there exists an integer a which when divided by 6 gives a remainder r.

Hence :

$\tt{ \purple{ \implies{a \: = 6q \: + \: r}}}$

Placing several values of r and equating, we obtain the above result.

Therefore we can define that there exists a '+ve' odd integer is of the form 6q+1 or 6q+3 or 6q+5