Q. 24.0( 83 Votes )

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Answer :

Let a be any odd 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 = 0,1,2,3,4,5because 0≤r<band the value of b is 6


So total form available will be 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5 ,6q + 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 but 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 but 4 is also divisible by 2, so 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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