If the product of all the unique positive divisors of a positive integ

Question

If the product of all the unique positive divisors of a positive integer  x  is  x\sqrt{x} , then the product of all the unique positive divisors of  x^4  is

A.  x^6 
B.  x^8 
C.  x^{16} 
D.  x^{18} 
E.  x^{36}

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IanStewart · Answer

Assuming x > 1 (if x = 1 every answer is right), then x always has two divisors, x and 1. Those two divisors alone give us a product of x, and if the product of all of x's divisors is x√x, then √x must be equal to the product of the remaining divisors of x. Those other divisors are all integers, so √x must be an integer, and if that's true, x is a perfect square. But if x is a perfect square, √x is automatically one divisor of x. So we've identified three necessary divisors of x: 1, √x and x, and these three divisors already multiply to x√x. So there can't be any other divisors.

If x has precisely three divisors, x can only be equal to p^2 for some prime p. The question asks about x^4, which equals p^8, which has nine divisors: 1, p, p^2, ... p^7, p^8. If we multiply those together, we'll just be adding all the exponents, so we'll get p^(1 + 2 + 3 + ... + 7 + 8) = p^36 = (p^2)^18, and since x = p^2, then x^18 is the answer.

If x has precisely three divisors, x can only be equal to p^2 for some prime p. The question asks about x^4, which equals p^8, which has nine divisors: 1, p, p^2, ... p^7, p^8. If we multiply those together, we'll just be adding all the exponents, so we'll get p^(1 + 2 + 3 + ... + 7 + 8) = p^36 = (p^2)^18, and since x = p^2, then x^18 is the answer.

sanjitscorps18 · Answer

Here  x  is itself the integer and all the divisors multiplied become  x\sqrt{x} 
The multiplication of possible divisors is usually the number itself, 1 and other divisors hence we have  x  =  1 * x * A , where A are the other divisors. Hence  x\sqrt{x}  =  1 * x * A 
=>  \sqrt{x}  =  A 
Since  \sqrt{x}  can only be an integer that cannot be further broken down it must be prime

Looking at the possible divisors of  x^4  we get
 x^4 ,  x^3\sqrt{x} ,  x^3 ,  x^2\sqrt{x} ,  x^2 ,  x\sqrt{x} ,  x ,  \sqrt{x}

Hence the multiplication becomes x^18

Option D

Regor60 · Answer

By examination one can pretty quickly see that 4 works as an example of X since:

1*2*4=8=4(4^.5)

And:

4^4=2^8, whose divisors are:

1,2,2^2...2^8

The sum of 1 to 8 is:

8×9/2 = 36

So the product of all the powers of 2 is:

2^36 = 4^18 = X^18

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If the product of all the unique positive divisors of a positive integ

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If the product of all the unique positive divisors of a positive integ

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