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

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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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New post 19 May 2015, 05:56
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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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New post 25 May 2015, 07:08
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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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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New post 19 May 2015, 08:04
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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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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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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New post 19 May 2015, 08:06
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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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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New post 19 May 2015, 08:14
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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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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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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New post 20 May 2015, 01:27
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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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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New post 22 May 2015, 22:41
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.

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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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New post 06 Oct 2015, 18:22
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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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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New post 07 Oct 2015, 05:42
1
1
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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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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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New post 07 Oct 2015, 12:33
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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Re: For prime numbers x and y, x^3*y^5=z^4. How many positive factors does  [#permalink]

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New post 11 Jan 2017, 07:38
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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