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Math Expert V
Joined: 02 Sep 2009
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For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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Question Stats: 40% (02:01) correct 60% (02:03) wrong based on 222 sessions

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

A. One
B. Two
C. Three
D. Four
E. Five

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Math Expert V
Joined: 02 Sep 2009
Posts: 55188
Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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1
8
Bunuel wrote:
For prime numbers x and y, x^3*y^5=z^4. How many positive factors does positive integer z have?

A. One
B. Two
C. Three
D. Four
E. Five

OFFICIAL SOLUTION:

Notice that z^4 is a perfect square, thus it must have odd number of factors. Since z^4 equals to x^3*y^5, then x^3*y^5 must also have odd number of factors.

Now, if x and y are different primes, then the number of factors of x^3*y^5 will be (3 + 1)(5 + 1) = 24, which is not odd. Therefore x and y MUST be the same prime.

In this case x^3*y^5 = x^3*x^5 = x^8 = z^4.
x^8 = z^4;
x^2 = z.

Since x is prime, then the number of factors of z is (2 + 1) = 3.

Answer: C.
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Math Expert V
Joined: 02 Aug 2009
Posts: 7681
Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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Bunuel wrote:
For prime numbers x and y, x^3*y^5=z^4. How many positive factors does positive integer z have?

A. One
B. Two
C. Three
D. Four
E. Five

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..
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Math Expert V
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Posts: 55188
Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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chetan2u wrote:
Bunuel wrote:
For prime numbers x and y, x^3*y^5=z^4. How many positive factors does positive integer z have?

A. One
B. Two
C. Three
D. Four
E. Five

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: $$x^3*y^5=z^4$$.
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Posts: 7681
Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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3
Bunuel wrote:
chetan2u wrote:
Bunuel wrote:
For prime numbers x and y, x^3*y^5=z^4. How many positive factors does positive integer z have?

A. One
B. Two
C. Three
D. Four
E. Five

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
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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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3
1
Bunuel wrote:
For prime numbers x and y, x^3*y^5=z^4. How many positive factors does positive integer z have?

A. One
B. Two
C. Three
D. Four
E. Five

Ans: C

Solution: given x^3*y^5=z^4; here if x and y are two different prime numbers then z = x^3/4 * y^5/4
in this case z can not be and integer.
now to make z = (x^3*y^5)^1/4 an integer it is required to have x=y then equation will become z=x^2 or z=y^2
hence total positive factors will be 3.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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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:

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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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Bunuel wrote:
Bunuel wrote:
For prime numbers x and y, x^3*y^5=z^4. How many positive factors does positive integer z have?

A. One
B. Two
C. Three
D. Four
E. Five

OFFICIAL SOLUTION:

Notice that z^4 is a perfect square, thus it must have odd number of factors. Since z^4 equals to x^3*y^5, then x^3*y^5 must also have odd number of factors.

Now, if x and y are different primes, then the number of factors of x^3*y^5 will be (3 + 1)(5 + 1) = 24, which is not odd. Therefore x and y MUST be the same prime.

In this case x^3*y^5 = x^3*x^5 = x^8 = z^4.
x^8 = z^4;
x^2 = z.

Since x is prime, then the number of factors of z is (2 + 1) = 3.

Answer: C.

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

Thanks.
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Math Expert V
Joined: 02 Sep 2009
Posts: 55188
Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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1
2
snfuenza wrote:
Bunuel wrote:
Bunuel wrote:
For prime numbers x and y, x^3*y^5=z^4. How many positive factors does positive integer z have?

A. One
B. Two
C. Three
D. Four
E. Five

OFFICIAL SOLUTION:

Notice that z^4 is a perfect square, thus it must have odd number of factors. Since z^4 equals to x^3*y^5, then x^3*y^5 must also have odd number of factors.

Now, if x and y are different primes, then the number of factors of x^3*y^5 will be (3 + 1)(5 + 1) = 24, which is not odd. Therefore x and y MUST be the same prime.

In this case x^3*y^5 = x^3*x^5 = x^8 = z^4.
x^8 = z^4;
x^2 = z.

Since x is prime, then the number of factors of z is (2 + 1) = 3.

Answer: C.

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.
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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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Bunuel wrote:
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! _________________
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GRE 1: Q169 V154 Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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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.

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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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