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If x, y, and z are integers greater than 1, and [#permalink]
 21 Feb 2010, 14:20
Question Stats:
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68% (03:38) wrong based on 167 sessions
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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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]
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]
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]
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]
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]
24 Feb 2010, 17:19
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Re: If x, y, and z are integers [#permalink]
01 Aug 2010, 16:31
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.
The correct answer is B.
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Re: If x, y, and z are integers [#permalink]
02 Aug 2010, 04:31
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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]
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]
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 Please explain in detail 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. Bunuel, can you please explain the answer for choice B
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Re: If x, y, and z are integers [#permalink]
28 Jun 2012, 02:53
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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 Please explain in detail 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. Bunuel, can you please explain the answer for choice B 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. Answer: B. Hope it's clear.
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Re: If x, y, and z are integers greater than 1, and [#permalink]
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.
Please let me know if I am thinking about this wrong
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Re: If x, y, and z are integers greater than 1, and [#permalink]
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.
Please let me know if I am thinking about this wrong 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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If x, y, and z are integers greater than 1 [#permalink]
08 Oct 2012, 04:09
If x, y, and z are integers greater than 1, and (327)(510)(z) = (58)(914)(xy), then what is the value of x?
(1) y is prime
(2) x is prime
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
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Re: If x, y, and z are integers greater than 1, and [#permalink]
22 Jan 2013, 01:21
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^y3*9^{13}*5^{10}*z = 5^8*9^{14}*x^y5^2*z = 3*x^yx^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. Answer: B
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If x,yand z are integers greater than 1,and [#permalink]
24 Feb 2013, 05:48
If x, yand 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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Re: If x,yand z are integers greater than 1,and [#permalink]
24 Feb 2013, 06:18
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Re: If x,yand z are integers greater than 1,and
[#permalink]
24 Feb 2013, 06:18
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