If Q the square of an odd positive integer and if 8Q^8 has four prime

Q8 must have 3 prime factors, and so does root over Q
can u plz explain this one

harivars

\(8*Q^8= 2^{3+d}*a^x*b^y*c^z \), where a, b and c are distinct odd prime numbers

As Q is square of odd integer, Q can't have 2 as its prime factor. d must be equal to 0.

so, \(Q^8 = a^x*b^y*c^z\)

\([Q^{1/2}]^{16}= a^x*b^y*c^z\)

\([Q^{1/2}]= a^{x/16}*b^{y/16}*c^{z/16}\)

\(Q^{1/2}\) has 3 prime factors a, b and c, which are same as that of Q^8.

8Q^8 has 4 prime factors. 8 has only one prime factor, ie 2, and Q does not have 2 as a prime factor as Q is odd as given.
Thus, the remaining 3 prime factors come from Q.
The number of prime factors doesn't change for square root. Does, 3 is the correct answer.

C is Correct

is an odd number, must be odd

Posted from my mobile device

Q is the square of an odd positive integer.
8Q^8 has four prime factors.
We want to know how many prime factors /sqrt{Q} has.

Let's start by thinking about 8Q^{8}.
--> We can write 8 as 2^{3} so we know 2 is one of the prime factors.
--> The other 3 prime factors must come from Q.
--> So Q either has 3 prime factors or 4 prime factors (if 2 | Q)

Let's now think about Q.
-->Let m be some odd positive integer.
--> Then there exists some even integer n >= 0 such that m = n + 1.
--> Then we can write Q = m^{2} = (n+1)^{2} = n^{2} + 2n + 1
--> We know that Q is an odd number now.
--> This means that 2 is not a factor of Q.
--> We can then say that Q has three distinct prime factors.

Now let's think about /sqrt{Q}.
--> It's helpful just to register that we know we're not dealing with any negative numbers.
--> /sqrt{Q} = /sqrt{(m)^{2})
--> Note that Q = m^{2} implies Q must have equally as many prime factors as m, since any additional factors would be composites resulting from multiplying some combinator of prime factor of m_1 against a combination of prime factors of m_2.
--> This means that /sqrt{Q} has equally as many prime factors as Q itself.
--> Then Q having three prime factors implies /sqrt{Q} has three prime factors.

In my previous post here, I showed why Q must have 3 prime factors. Here I'll show for a general case why Q having 3 prime factors implies Q^(1/2) must have 3 prime factors.

Suppose we have some integer z > 1.

As we always do, we can decompose z into the product of any grouping of its factors: z = f_1 * f_2 * f_3 * f_4 ... f_n

Without the loss of generality, we can further break up the factors of z into powers of its unique prime factor, e.g.: z = p_1^a * p_2^b * p_3^c ...

What happens when we square z? We just get z * z = (p_1^a * p_2^b * p_3^c ...) * (p_1^a * p_2^b * p_3^c ...) = (p_1^2a * p_2^2b * p_3^2c ...)

So taking powers of z just changes the exponents on its prime factors -- it doesn't change the count of unique prime factors p_1 ... p_n. You can use induction to show this for any power over 1, which is kinda neat.

Because we know Q is the square of an integer, we can safely apply the logic backwards, and just halve the exponents on its prime factors.

----- Example ----

Suppose z is 12^2 = 144

We can break up z into 12 * 12 = 3 * 4 * 3 * 4 = 3^2 * 2^4; it has just two distinct prime factors.

If we take the root of z, that's just 12, which is simply 3^1 * 2^2. Note that we just cut the exponents in half.

Asked: If Q the square of an odd positive integer and if 8Q^8 has four prime factors, then how many prime factors does √Q have?

Q = (2k+1)^2
2^3(2k+1)^8 has 4 prime factors ; 2k+1 has 3 prime factors; √Q = 2k + 1 has 3 prime factors

IMO C

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