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Question

N is a natural number, then how many values of N are possible such that 6N3+3N2+N+24N
is also a Natural Number?

A
6
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B
7
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C
8
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D
9
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Solution

The correct option is C 8
Suppose

6N3+3N2+N+24N=M=Natural Number

So, M=6N2+3N+1+24N

Thus to be M as an natural number

and 24N should be Natural number

Thus N should be factor of 24.

So, 24=23×31

Factors = 4 x 2 = 8

8 values can be taken by N

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