For prime numbers x and y, x^3*y^5=z^4. How many positive factors does

hi,
does the Q mean (x^3)*(y^5)=z^4, because in that way positive integer z is not possible...
it should be .x^(3*y^5)=z^4.. if this is what it means answer should be 4..

The equation is correct: \(x^3*y^5=z^4\).

thanks!
in that case only possibility is when x=y..
so x^8=z^4... or (x^2)^4=z^4... so z=x^2.. so z has three factors.... ans C

Hi All,

This question can be solved by TESTing VALUES, but it comes with a 'design issue' that might cause you some trouble if you make the work more complicated than it needs to be.....

We're told that X and Y are PRIME NUMBERS and we're told that (X^3)(Y^5) = Z^4.

Since this equation is rather "ugly", I would pick the easiest prime that I could think of: 2 (and I'd use it for BOTH X and Y).....

Thus, (2^3)(2^5) = (8)(32) = 256 = Z^4

256 = 2^8 = 4^4 = Z^4

So, Z = 4

We're asked for the number of positive factors that Z has...

The factors of 4 are 1, 2 and 4, so there are 3 total factors.

Final Answer:
GMAT assassins aren't born, they're made,
Rich

Bunuel, can you elaborate on why z^4 must have an odd number of factors, because it is a combination of prime numbers?

Thanks.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

BACK TO YOUR QUESTION:
In case of a perfect square, say for \(6^2 = 2^2*3^2\), the powers of primes will be always even, so when counting the number of factors we'd have \((even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd\).

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

Hope it helps.

Thanks Bunuel, you're the best!

This is a great Question.
Here is what i did in this one =>
Given x and y are primes.
And z is a postive integer
So z^4 will be a perfect 4th power.
So the exponent of all the primes must be a multiple of 4.
But wait z=x^3*y^5 and x and y are prime numbers.
The only way the primes will have the exponent which will be a multiple of 4 is when x=y
Putting x=y=> z^4=x^8 => z=x^2=> 3 factors/

Hence C.

Bunuel has a much better solution, but here was my approach.
This is a PS question (so the answer that's true for one case, will be true for a general case). So we need to find an answer, and that will be THE ANSWER (this is why number picking works so well in PS). Anyway, with that said

x^5 y^3 = z^4
Let's consider x = y (because, as explained before, if we find an answer for x = y, we can be sure that's the answer)
Why did I do this? Coz 5 + 3 = 8 = 4*2
so x^8 = z^4
so z = x^2 (where x is prime)
so z can have 3 factors : x^0, x^1 and x^2

Since in gmat, answer must be true for this kind of question for all scenarios, can we not easily solve it with number plucking in a much faster way? Or there is any downside to number plucking?

E.g, if we pluck x=y=2, to satisfy the equation, then,
2^8=(2^2)^4,
then n=2^2

Total factors = 3

This particular problem is testing your knowledge on factors of perfect square.

Number picking worked luckily for you because you took x = y, what if you had picked x = 2, and y = 3? You would've eventually had to come to terms that x = y and by that time you could have wasted your precious time on test. Once you know, x = y, you don't even need to test numbers, \(x^8 = z^4 => x^2 = z\)

Picking numbers isn't a wise strategy for this question- unless of course, after picking numbers, if you quickly identify that x has to be equal to y.