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

Let take a as any positive integer and b = 6. a > b

Then using Euclid’s algorithm, we get a = 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 < b and the value of b is 6


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


6q + 0 6 is divisible by 2 so it is an even number.


6q + 1 6 is divisible by 2 but 1 is not divisible by 2 so it is an odd number.


6q + 2 6 is divisible by 2 and 2 is also divisible by 2 so it is an even number.


6q + 3 6 is divisible by 2 but 3 is not divisible by 2 so it is an odd number.


6q + 4 6 is divisible by 2 and 4 is also divisible by 2 it is an even number.


6q + 5 6 is divisible by 2 but 5 is not divisible by 2 so it is an odd number.


So odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5.


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