Question

How to 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 take x as any positive integer and y = 6.

Then using Euclid’s algorithm we get  x = 6q + r  here r is remainder and value of q is more than or equal to 0  and r = 0, 1, 2, 3, 4, 5 because 0 ≤r < y  and the value of y is 6.

So total possible forms  will  6q+0 , 6q+1 , 6q+2,6q+3,6q+4,6q+5 

6q+0   6 is divisible by 2; thus it is a even number

6q+1   6 is divisible by 2 but 1 is not divisible by 2; thus it is a odd number

6q+2  6 is divisible by 2   and 2 is also divisible by 2; thus it is a even number

6q+3  6 is divisible by 2 but 3 is not divisible by 2; thus it is a odd number

6q+4  6 is divisible by 2 and 4 is also divisible by 2; thus it is a even number

6q+5  6 is divisible by 2 but 5 is not divisible by 2; thus it is a odd number

So all odd integer will be in form of 6q + 1, or 6q + 3, or 6q + 5

Answer 2

Answer:

hope it helps you

please markmy answer as brainlist

Step-by-step explanation:

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.

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

Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly,

6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or

6q + 3, or 6q + 5