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Question

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

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Solution

## Let's take 'a' as any positive integer and b = 6. 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, any positive odd integer will be of the form 6q + 1, or 6q + 3, or 6q + 5.

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