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Question

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


Solution

N=(a+ib)3107i=a3+3a2bi+3ab2i2+b3i3107i=(a33ab2)+i(3a2bb3107)
Since N is positive integer, its imaginary part equals zero.
b(3a2b2)=107
Since 107 is prime, hence it has exactly two divisors which are 1 and 107 
Hence, b=1 and 3a2b2=107 
If 3a2b2=1 and b=107, then a is not an integer. 
3a2b2=107
3a2=108a2=36
a=6   (Neglecting a=6)
N=216361=198 

Mathematics

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