Question

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

Answer 1

we have a=bq+r
here b is greater than r
then r is greater than or equal to 0
in the given equation the value of b is 6
therefore the possible values of r are 0,1,2,3,4 and 5
if we take r= 1 then

a=6q+1=2 (3q)+1
now here you assume 3q as q
then we ger 3q 2q+1 which is the general form of an odd no.

if r=0
then a=6q+0=6q
2 (3q)
if we assume 3q as m then it is a=2m which is the general form of an even integer

if r is 2 then
a=6q+2
=2 (3q+1)
assume 3 q+1 as m
then you get it as2m it is the general form of an even integer

now also try the other values of the r hope you understand

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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