N is a natural number. How many values of N exist, such that N2+24N+21 has exactly three factors?

Options:

A . 0

B . 1

C . 2

D . 3

E . >3

Answer: Option C : C Solution:Option c For N2+24N+21 to have exactly three factors, it must be square of a prime number. Let N2+24N+21=a2 where a is a prime number. (N+12)2−123=a2 (N+12+a)(N+12-a) = 123 Either N+12+a =123 and N+12-a = 1 Or N+12+a = 41 and N+12-a=3 In the first case N = 50 and a = 61. In the second case N =10 and a =19 In either case N2+24N+21 is the square of a prime number. So two such values exist.

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