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# If x, y, and z are integers greater than 1, and

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If x, y, and z are integers greater than 1, and  [#permalink]

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Updated on: 08 Oct 2012, 04:18
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Question Stats:

44% (02:28) correct 56% (02:24) wrong based on 1525 sessions

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If x, y, and z are integers greater than 1, and $$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$, then what is the value of x?

(1) y is prime.
(2) x is prime.

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Originally posted by jeeteshsingh on 21 Feb 2010, 14:20.
Last edited by Bunuel on 08 Oct 2012, 04:18, edited 2 times in total.
Edited the question.
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Re: If x, y, and z are integers  [#permalink]

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28 Jun 2012, 02:53
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11
riteshgupta wrote:
Bunuel wrote:
jeeteshsingh wrote:
If x, y, and z are integers greater than 1, and $$3^{27}*5^{10}*z = 5^8*9^{14}*x*y$$, then what is the value of x?

(1) y is prime

(2) x is prime

SOURCE: Manhattan tests

Spoiler: :: OA
B

If the answer is really B, then I think question should be: $$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$. If I'm right, then it's B indeed.

In it's current form the answer is C as explained.

You mean what would be the solution if it were x^y instead of xy?

If x, y, and z are integers greater than 1, and $$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$, then what is the value of x?

$$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$ --> $$3^{27}*5^{2}*z =3^{28}*x^y$$ --> $$5^{2}*z = 3*x^y$$ --> $$\frac{x^y}{z}=\frac{5^2}{3}$$, so $$x^y$$ is a multiple of 25 and $$z$$ is a multiple of 3.

(1) y is prime. We can have that $$x=5$$, $$y=2=prime$$ and $$z=3$$ OR $$x=10$$, $$y=2=prime$$ and $$z=12$$... Not sufficient.

(2) x is prime. Since $$x^y$$ is a multiple of $$5^2$$ and $$x$$ is a prime, then $$x=5$$. Sufficient.

Hope it's clear.
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Re: If x, y, and z are integers  [#permalink]

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22 Feb 2010, 11:19
(327)(510)(z) = (58)(914)(xy)

3 * 109 * 3 * 17 * 2 * 5 * z = 2 * 29 * 2 * 457 * x * y (all the numbers are prime)

x = (3^2 * 5 * 17 * 109 * z) / (29 * 2 * 457 * y) (cancelling 2 on numerator and denominator)

1) y is prime
Not sufficient
y = 3, z = 29*2*457 * n (where n is an intereger ) and values of x could vary depending on n or depending on y(3,5,109, 457 etc)
(2) x is prime
Not sufficient
y = 9 * 5 * 17,z = 29*2*457 , x = 109
y = 9 * 5 * 109,z = 29*2*457 , x = 17

combining .. there is no value of x that will satify the equation as if y is prime, then y could be any of the values of 3,5, 17, 109 or a different number that is a factor of z. and if z is an interger, x cannot be prime.

E ( not sure if E is the correct answer as there is no valid value of x that satisfies the conditions)
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Re: If x, y, and z are integers  [#permalink]

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24 Feb 2010, 16:29
IMO C

Reduced equation becomes

$$\frac{25*z}{3^{15}*x}$$ = y

now if y is prime

z can be 3^15 and x = 5 or z = 2* 3^15 and x = 10 thus not sufficient

now if x is prime same explanation above.

now take both x and y prime, in this case x can be only 5 and z = 3^15

thus it should be C
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Re: If x, y, and z are integers  [#permalink]

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24 Feb 2010, 16:36
Here's the solution for the re-posted problem
If x, y, and z are integers greater than 1, and 3^27 * 5^10 * z =5^8 * 9^14 * x * y , then what is the value of x?
3^27 * 5^10 * z =5^8 * 9^14 * x * y
3^27 * 5^ 10 * z = 5^ 8 * 3^28 * x * y
5^2 * z = 3 * x * y
x = (5^2 * z) / 3y

(1) y is prime
y can be a factor of z, and z can be a factor of y and there could be a lot of possibilities of x
y = 3, z = 27, x = 75
y = 5, z = 3, x = 5
Not sufficient
(2) x is prime
z has to be a multiple of 3 and y has to be a multiple of 5
z= 3*a (where a is prime not equal to 5), y = 25, then x=a(many values for a)
z=3, y = 5, x=5
Not sufficient

combining .. x and y both are prime .. and as 5^2 is in the numerator, y has to be 5 and z has be to 3 or else x cannot be prime.
y=5, z=3, x=5

C
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Re: If x, y, and z are integers  [#permalink]

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24 Feb 2010, 16:45
After simplifying the given question, we end up with 5*5*z/(3*y) = x.

Stmt 1 - y can take a whole lot of prime numbers from 2 to infinity (not the car!!!) and until we know z, we can't comment on value of x.

Stmt 2 - x itself can take a whole lot of prime numbers from 2 to infinity (again, not the car!!!) and we would have to struggle to get appropriate values for z and y, and there can very many that match the criteria.

Combining both stmts - as long as z is some composite, that can be expressed as a multiple of 3, x (any prime number choice) and y (prime number choice), it is good, and there are whole lot of possibilities for x as well.

I am not very sure, how do we get to B as the OA. In my opinion, it must be E. Math experts and Math God Bunuel might wish to throw some light!!!!
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Re: If x, y, and z are integers  [#permalink]

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24 Feb 2010, 17:19
4
1
jeeteshsingh wrote:
If x, y, and z are integers greater than 1, and $$3^{27}*5^{10}*z = 5^8*9^{14}*x*y$$, then what is the value of x?

(1) y is prime

(2) x is prime

SOURCE: Manhattan tests

Spoiler: :: OA
B

If the answer is really B, then I think question should be: $$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$. If I'm right, then it's B indeed.

In it's current form the answer is C as explained.
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Re: If x, y, and z are integers  [#permalink]

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01 Aug 2010, 16:31
2
I had this question in the test today too. I went with E but QA is B. I didnt understand the explanation. I think QA is wrong here, 2 is a prime too.

---

The best way to answer this question is to use the rules of exponents to simplify the question stem, then analyze each statement based on the simplified equation.

(327)(510)(z) = (58)(914)(xy) Simplify the 914
(327)(510)(z) = (58)(328)(xy) Divide both sides by common terms 58, 327
(52)(z) = 3xy

(1) INSUFFICIENT: Analyzing the simplified equation above, we can conclude that z must have a factor of 3 to balance the 3 on the right side of the equation. Similarly, x must have at least one factor of 5. Statement (1) says that y is prime, which does no tell us how many fives are contained in x and z.

For example, it is possible that x = 5, y = 2, and z = 3:
52 · 3 = 3 · 52

It is also possible that x = 25, y = 2, and z = 75:
52 · 75 = 3 · 252
52 · 52 · 3 = 3 · 252

(2) SUFFICIENT: Analyzing the simplified equation above, we can conclude that x must have a factor of 5 to balance out the 52 on the left side. Since statement (2) says that x is prime, x cannot have any other factors, so x = 5. Therefore statement (2) is sufficient.

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Re: If x, y, and z are integers  [#permalink]

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02 Aug 2010, 04:31
1
what if y=25 , z=6 and x=2 ?
if y =5, x=5 and z=3 ?

for statement 2 both holds true..so B alone is not sufficient.
either OA is wrong or Question is wrong.
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Re: If x, y, and z are integers  [#permalink]

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02 Aug 2010, 04:50
The answer has to be C..

3^27 * 5^10 * z =5^8 * 9^14 * x * y
3^27 * 5^ 10 * z = 5^ 8 * 3^28 * x * y
5^2 * z = 3 * x * y
x = (5^2 * z) / 3y

after this..either of the statements ie..

1) X is prime

y and z can take a whole lot of values..

2) y is prime.

x and z can again take many values..prime or not prime..

hence only when we know that both x and y are prime
can we reach the answer that z=3, x=5,y=5
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Re: If x, y, and z are integers  [#permalink]

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27 Jun 2012, 09:36
Bunuel wrote:
jeeteshsingh wrote:
If x, y, and z are integers greater than 1, and $$3^{27}*5^{10}*z = 5^8*9^{14}*x*y$$, then what is the value of x?

(1) y is prime

(2) x is prime

SOURCE: Manhattan tests

Spoiler: :: OA
B

If the answer is really B, then I think question should be: $$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$. If I'm right, then it's B indeed.

In it's current form the answer is C as explained.

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Re: If x, y, and z are integers greater than 1, and  [#permalink]

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17 Jul 2012, 08:20
think the answer is B even as written?

Formula simplifies to 25z/3y = x or 25z/3x=y

1: y is prime; since 25z/3y yields an integer, does it not hold that 25/y = integer since 3 is not a factor of 25. If y is prime, it has to be 5 since 5^2 are the prime factors of 25. However, in order to know X, we need to know what z might be, and Z could be any number with 3 as a prime factor. Not sufficient.

2: x is prime; since 25z/3x yields an integer, 25/x is also an integer since 3 is not a factor of 25. Holding the same logic as in 1; x must be 5. Sufficient.

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Re: If x, y, and z are integers greater than 1, and  [#permalink]

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17 Jul 2012, 16:41
JohnGalt44 wrote:
think the answer is B even as written?

Formula simplifies to 25z/3y = x or 25z/3x=y

1: y is prime; since 25z/3y yields an integer, does it not hold that 25/y = integer since 3 is not a factor of 25. If y is prime, it has to be 5 since 5^2 are the prime factors of 25. However, in order to know X, we need to know what z might be, and Z could be any number with 3 as a prime factor. Not sufficient.

2: x is prime; since 25z/3x yields an integer, 25/x is also an integer since 3 is not a factor of 25. Holding the same logic as in 1; x must be 5. Sufficient.

Yes you are thinking about it wrongly.
Why does 25/x have to be an integer? 3 is not a factor of 25, but it can be a factor of z, the other term in the numerator.
Hence, if you let y=5^2, then z=3x, and x=any prime you want it to be. Not sufficient

However taking (1)+(2), both x and y have to be prime and since you need to make 25 using 3*x*y, only way to do it is using x=y=5 and z=3. C.

Please give kudos if you like.

Also, can OP please change answer in spoiler to the correct one?
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Re: If x, y, and z are integers greater than 1, and  [#permalink]

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22 Jan 2013, 01:21
1
jeeteshsingh wrote:
If x, y, and z are integers greater than 1, and $$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$, then what is the value of x?

(1) y is prime.
(2) x is prime.

$$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$
$$3*9^{13}*5^{10}*z = 5^8*9^{14}*x^y$$
$$5^2*z = 3*x^y$$
$$x^y = \frac{5^2*z}{3}$$

It's obvious that z has a factor of 3 to cancel out the denominator and x has 5 as a factor...

1. y is prime

Let y = 3: $$5^3 = \frac{5^2*(5*3)}{3}$$ Thus, x=5
Let y = 3 and x contain 3: $$5^{3}*3^3 = \frac{5^2*(5*3^4)}{3}$$ Thus, x=15

INSUFFICIENT.

2. x is prime.. Well it's obvious that x has 5 as a factor.. If it's prime then x = 5*1 = 5
SUFFICIENT.

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Re: If x, y, and z are integers  [#permalink]

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18 Jul 2013, 15:05
Bunuel wrote:

You mean what would be the solution if it were x^y instead of xy?

If x, y, and z are integers greater than 1, and $$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$, then what is the value of x?

$$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$ --> $$3^{27}*5^{2}*z =3^{28}*x^y$$ --> $$5^{2}*z = 3*x^y$$ --> $$\frac{x^y}{z}=\frac{5^2}{3}$$, so $$x^y$$ is a multiple of 25 and $$z$$ is a multiple of 3.

(1) y is prime. We can have that $$x=5$$, $$y=2=prime$$ and $$z=3$$ OR $$x=10$$, $$y=2=prime$$ and $$z=12$$... Not sufficient.

(2) x is prime. Since $$x^y$$ is a multiple of $$5^2$$ and $$x$$ is a prime, then $$x=5$$. Sufficient.

Hope it's clear.

Bunuel, if the statement (1) were- Z (instead of Y) is prime then the answer should be 'D'? Thanks!
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Re: If x, y, and z are integers  [#permalink]

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18 Jul 2013, 15:17
mneeti wrote:
Bunuel wrote:

You mean what would be the solution if it were x^y instead of xy?

If x, y, and z are integers greater than 1, and $$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$, then what is the value of x?

$$3^{27}*5^{10}*z = 5^8*9^{14}*x^y$$ --> $$3^{27}*5^{2}*z =3^{28}*x^y$$ --> $$5^{2}*z = 3*x^y$$ --> $$\frac{x^y}{z}=\frac{5^2}{3}$$, so $$x^y$$ is a multiple of 25 and $$z$$ is a multiple of 3.

(1) y is prime. We can have that $$x=5$$, $$y=2=prime$$ and $$z=3$$ OR $$x=10$$, $$y=2=prime$$ and $$z=12$$... Not sufficient.

(2) x is prime. Since $$x^y$$ is a multiple of $$5^2$$ and $$x$$ is a prime, then $$x=5$$. Sufficient.

Hope it's clear.

Bunuel, if the statement (1) were- Z (instead of Y) is prime then the answer should be 'D'? Thanks!

Yes, that's correct.
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Re: If x, y, and z are integers greater than 1, and  [#permalink]

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27 Oct 2013, 16:34
Came across this question just now, it's a good one to study exponents:
if 3^27.5^10.z = 9^14.5^8.x^y what is the value of x?
the statement can be further expanded becoming: 3^27.5^10.z = 3^28.5^8.x^y

1) y is prime, we can be tempted to say y = 2, this is true in the case where z = 3 then x becomes 5 but what if z = 15 =(5)(3) then x^y = ((5)(3))^3 where x = (5)(3) and y = 3 where y is still prime. So insufficient.

2) x is prime, this leaves no doubt that x must equal 5.
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Re: If x, y, and z are integers greater than 1, and  [#permalink]

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16 Mar 2014, 04:35
But it says x,y,z are integers so they should be a single digit .. why are we taking them as double digits .. ?
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Re: If x, y, and z are integers greater than 1, and  [#permalink]

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16 Mar 2014, 05:13
arunmunich wrote:
But it says x,y,z are integers so they should be a single digit .. why are we taking them as double digits .. ?

Why should x, y, and z be single digit integers??? Which solution are you referring to???
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Re: If x, y, and z are integers greater than 1, and  [#permalink]

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30 Mar 2014, 12:46
I think it goes this way...

In order for the 2 parts to be equal we need to have the power of 3 in one side equal to the power of 3 in the other one.Same goes for 5.So in the left one the power of 3 is 27 and in the other one is 28.So we need a 3.In the left one the power of 5 is 10 and in the right is 8.

If we know for sure that x is a prime then it z must be 3(because it cannot be 5^-2) and x has to be the 5 missing..!!!!
Re: If x, y, and z are integers greater than 1, and   [#permalink] 30 Mar 2014, 12:46

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