InterviewSolution
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InterviewSolution
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ST S . S छः2. दर्शाइए कि कोई भी धनात्मक विषम पूर्णाक 60+1 या 6८ +3 या 6/+5 के रूप का होता है,जहाँ / कोई पूर्णाक है। |
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Answer» Let take a as any positive integer and b = 6.Then using Euclids 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 6So total possible forms will 6q+0 , 6q+1 , 6q+2,6q+3,6q+4,6q+56q+0 6 is divisible by 2 so it is a even number6q+1 6 is divisible by 2 but 1 is not divisible by 2 so it is a odd number6q+2 6 is divisible by 2 and 2 is also divisible by 2 so it is a even number6q+3 6 is divisible by 2 but 3 is not divisible by 2 so it is a odd number6q+4 6 is divisible by 2 and 4 is also divisible by 2 it is a even number6q+5 6 is divisible by 2 but 5 is not divisible by 2 so it is a odd numberSo odd numbers will in form of 6q + 1, or 6q + 3, or 6q + 5 Like my answer of you find it useful! |
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