If P is a prime number and Q is a positive integer, how many factors does P^2*Q have?

(1) The lowest number that has both P^2 and Q^3 as its factors is 5400.

This means that 5,400 is the least common multiple of P^2 (square of a prime) and Q^3 (perfect cube), so both P^2 and Q^3 are factors of 5,400. Factorize: \(5400 = 2^3*3^3*5^2\). P^2, which is a square of a prime, must be 5^2 (so P must be 5) because in any other case (say if P^2 is 2^2 or 3^2), the remaining multiple will not be a perfect square (Q^3). Therefore, \(P^2 = 5^2\) (P = 5) and \(Q^3 = 2^3*3^3\) (Q = 6).

\(P^2*Q = 5^2*6 = 5^2*2*3\). The number of factors \(= (2 + 1)(1 + 1)(1 + 1) = 12\).

Sufficient.

(2) P and Q have only one common factor. This means that P and Q are co-prime, their only common factor is 1. This is clearly insufficient: for example, P could be 2 and Q could be 3, 3^2, 3^3, ...

Answer: A.

Hope it's clear.
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Hi Bunuel,

Why are we assuming P^2 and Q^3 are the only factors apart from 1 for the number 5400?

I considered multiple possibilities for Q [2,3,6,1] so went with E. I think I am missing something in the reasoning. Can you help?

Tx.
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How can P^2 and Q^3 be the only factors of 5,400? A number to have four factors should be of the form prime1*prime2, in this case its factors would be 1, p1, p2, and p1*p2.

If you say that Q can 1, 2, or 3, then what would P and would 5,400 be the LCM of P^2 an Q^3?
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Are 2 prime numbers coprime? 2 and 7, for example. Thanks
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Oh, I found the answer. Thanks!

Two integers a and b are said to be coprime if the only positive integer that evenly divides both of them is 1. That is, the only common positive factor of the two numbers is 1. So, according to the definition any two distinct primes are coprime.
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Two numbers are co-prime if they do not share any common factor but 1. So, any two different primes are co-prime. For example, 2 and 7 do not share any common factor but 1.
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Thanks got my error. The only possible combination is 6^3*5^2 because, the numbers have to be coprime [The composite number cannot contain the prime number here], so the LCM is effectively product of the two. Should have noticed.
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Hi Bunuel,
Regarding 2nd statement ,since P and Q are coprime,I guess the number of factors will not change because if we take P=2 and Q=3 then number of factors for P^2Q will be (2+1)(1+1)=6 and if we take P=2 and Q=7 then also number of factors will remain same.Kindly let me know if i am missing anything.

First of all, please re-read this:
(2) P and Q have only one common factor. This means that P and Q are co-prime, their only common factor is 1. This is clearly insufficient: for example, P could be 2 and Q could be 3, 3^2, 3^3, ...

Next, try to pick P and Q so that they are co-prime but so that both are not primes.

This should help.
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Oh ok..Thanks if i pick 7 and 9 ,no. of factors will vary.

If P is a prime number and Q is a positive integer, how many factors does P^2*Q have?

Statement 1:-
(1) The lowest number that has both P^2 and Q^3 as its factors is 5400.

This means that 5,400 is the LCM of P^2 (square of a prime) and Q^3 (perfect cube).
So both P^2 and Q^3 are factors of 5,400.
Factorize: \(5400 = 2^3*3^3*5^2\). P^2, which is a square of a prime, must be 5^2 (so P must be 5)
Therefore, \(P^2 = 5^2\) (P = 5)
And \(Q^3 = 2^3*3^3\) (Q = 6).

\(P^2*Q = 5^2*6 = 5^2*2*3\). The number of factors \(= (2 + 1)(1 + 1)(1 + 1) = 12\).

Sufficient.

Statement 2 :-
P and Q have only one common factor. This means that P and Q are co-prime,.
For example, P could be 2. Q could be 3, 3^2, 3^3, ... There is no data provided for Q.
This is clearly insufficient.

Answer: A.

Hi Bunuel

Are 5 and 12 are co-primes ?

Thanks

Yes.

The factors of 5 are 1 and 5.
The factors of 12 are 1, 2, 3, 4, 6 and 12.

As you can see the only factor they share is 1, so 5 and 12 are co-prime.
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Hi Bunuel,

Could you please explain this again or may be little deeper with some example ? I also picked up E as I am also having the same doubt which srinjoy1990 has.

Thanks a lot.