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A number N when factorised can be written as N = P1^4 x P2^3 x P3^7. Find the number or perfect squares which are factors of N. (the three prime numbers P1, P2, P3 > 2)?
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A number N when factorised can be written as N = P1^4 x P2^3 x P3^7. F...
Approach:

To find the number of perfect squares which are factors of N, we need to consider the powers of prime factors of N. A perfect square has an even power of its prime factors. Therefore, we need to find the number of combinations of even powers of P1, P2, and P3 that can be formed.

Solution:

Given, N = P1^4 x P2^3 x P3^7

Let the power of P1 be even, i.e., 0, 2, or 4
Let the power of P2 be even, i.e., 0 or 2
Let the power of P3 be even, i.e., 0, 2, 4, 6 or 7

Therefore, the possible combinations of even powers of P1, P2, and P3 are:

For P1, there are 3 possible even powers: 0, 2, and 4
For P2, there are 2 possible even powers: 0 and 2
For P3, there are 3 possible even powers: 0, 2, and 4

Hence, the total number of combinations of even powers of P1, P2, and P3 is:

3 x 2 x 3 = 18

Therefore, there are 18 perfect squares that are factors of N.

Answer:

The number of perfect squares which are factors of N is 18.
Community Answer
A number N when factorised can be written as N = P1^4 x P2^3 x P3^7. F...
N = P1^a + P2^b + P3^c

Allowed values of a,b,c for N to be a perfect square :-
a = { 0,2,4 }
b = { 0,2 }
c = { 0,2,4,6 }

combinations of these sets values.
Ans = 3*2*4 = 24
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A number N when factorised can be written as N = P1^4 x P2^3 x P3^7. Find the number or perfect squares which are factors of N. (the three prime numbers P1, P2, P3 > 2)?
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