Question
N is a natural number. How many values of N exist, such that N2+24N+21 has exactly three factors?
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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:
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.
Was this answer helpful ?
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