Question

N is a natural number. How many values of N exist, such that $N^2+24N+21$ has exactly three factors?

• A. 1
• B. 2
• C. 3
• D. >3

Answer 1

The correct option is B 2

Solution:Option c

For $N^2+24N+21$ to have exactly three factors, it must be square of a prime number.

Let $N^2+24N+21=a^2$ where a is a prime number.

$(N+12{)}^2-123=a^2$

(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 $N^2+24N+21$ is the square of a prime number. So two such values exist.