Question

N is a natural number, then how many values of N are possible such that $\frac{6N^3+3N^2+N+24}{N}$
is also a Natural Number?

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

The correct option is C 8

Suppose

$\frac{6N^3+3N^2+N+24}{N}=M=NaturalNumber$

$So,M=6N^2+3N+1+\frac{24}{N}$

Thus to be M as an natural number

$and\frac{24}{N}shouldbeNaturalnumber$

Thus N should be factor of 24.

$So,24=2^3\times 3^1$

Factors = 4 x 2 = 8

8 values can be taken by N