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Question

The number of terms with integral coefficient in the expansion of (27)16+1032x600 is

A
601
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B
301
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C
300
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D
302
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Solution

The correct option is B 301
(2716+32110x)600
=(312+212)600
Total number of integral terms will be
=600L.C.M(2,2)+1
=6002+1
=301

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