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Question

Let c=(a+ib)347i, where a,b,cZ+ and i=1. Then the value of a+b+c is

A
52
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B
49
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C
57
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D
63
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Solution

The correct option is C 57
c+47i=(a33ab2)+(3a2bb3)i
c=a33ab2 and 47=3a2bb3

47=b(3a2b2)
Since a,bZ+,
b is a divisor of 47.
b=1 or b=47

If b=47
3a2=1+472
Since, 1+472 is not divisible by 3, b cannot be 47.
b=1

a=4 (aZ+)
c=a33ab2=52
a+b+c=57

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