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Question

If a and b are positive integers such that N=(a+ib)3107i is a positive integer then N6 is

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Solution

N=(a+ib)3107i
=a3ib3+3a2ib3ab2107i
=(b3+3a2b107)i+a33ab2
Since, N is a positive integer so,
Im(z)=0
b3+3a2b107=0b(3a2b2)=107
b=1 and 3a2b2=107a=6
N=a33ab2=198
N6=33

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