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.
Submit Comment/FeedBack