zero

A number more than 1  always have  two or more than two factors.

1 and number itself is always among the factors.

lets take the example of some natural numbers having exactly three factors.

4 = 1,2,4

9= 1,3,9

25=1,5,25

49=1,7,49

so we can observe that for having exactly three factors the number should be perfect square of the prime number.

but N^2+24N+21 is not even a perfect square. So no value of  N can satisfy this .

so answer is zero...

Correct me if i am wrong.....

Solution

Question: How many values of N exist, such that (N^2 + 24 N +21) has exactly 3 factors?N is a natural no.

Solution:

We know that any square of a prime number, that is, a number of the form, (p^2) will always have Exactly 3 divisors, and they are, 1, p and (p^2). Hence we have to find a prime number or prime numbers (if it exists), say, 'a', such that:

N^2 + 24N + 21  = a^2

or, N^2 + 24N + {21 - (a^2)} = 0

Treating the above equation as a quadratic equation in N, and using Shridhacharya's Formula we have:

N = [ - 24 +- square root {(24^2) - 4(1){21 - (a^2)}]/{(2)(1)} ------------- (1)

Note: Shridharacharya's Method for finding the roots of a Quadratic Equation of the form Ax^2 + Bx + C = 0, is:

x = [ - B +- square root {(B^2) - 4(A)(C)}/{(2)(A)}]

On Simplification of expression (1), we have:

N = [ - 24 +- 2{square root {(a^2) + 123}]/2

or, N = - 12 +- square root{(a^2) + 123} ------------- (2)

Since it is given that N is a Natural Number, that is, a Positive Integer, that is, a Real Number, the Discriminant of the expression (2), that is, {(a^2) + 123} must be a perfect square.

Now, it cannot be equal to zero, since in that case, we will get: a^2 = - 123 and we know that the square of any quantity cannot be negative. Hence it has to be a perfect square greater than zero.

Now, let us suppose that (a^2) + 123 = (b^2) for some integer b.

Then, we have: (b^2) - (a^2) = 123

or, (b + a)(b - a) = 123

Now, the left hand side is a product of two factors, so we express the right hand side too as a product of two factors in all possible ways.

(b + a)(b - a) = 123 x 1 ------------- (3)

(b + a)(b - a) = 41 x 3 ------------- (4)

(b + a)(b - a) = 3 x 41 ------------- (5)

(b + a)(b - a) = 1 x 123 ------------- (6)

Now, let us analyse equation (3) first in the following manner:

When b + a = 123, b - a = 1

So, on solving, we get: b = 62 and a = 61 and accordingly, if we substitute the value of 'a' in equation (2), we get the values of N as: -74 and 50, but since N is a Natural Number, N cannot be - 74. Hence N = 50.

Now, Similarly, we can solve for 'a' and 'b' from equations (4), (5) and (6) and obtain the values of 'a' and 'b' as:

From equation (4): a = 19, b = 22, and therefore, we get N = -34 and 10

From equation (5): a = - 19, b = 22, and therefore, we get N = -34 and 10

From equation (6): a = - 61, b = 62, and therefore, we get N = -74 and 50

From the above values of N, we reject the negative ones, since N is a Natural Number and we consider the positive ones, thus getting N = 10 and 50, for which the corresponding values of the expression (N^2 + 24N + 21) are: 361, which is equal to (19^2) and 3721, which is equal to (61^2) and each of these have exactly 3 factors.

Hence there are exactly "Two" values of N (10 and 50) for which the expression (N^2 + 24N + 21) has exactly 3 factors.

Thank You.

Ravi Raja

Number System

Thanks a lot 4 ur kind explanation.

Answer

Try n=10.  Result is 361=19^2.