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If x, y, and z are integers greater than 1, and (3^27)(35^10
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Updated on: 31 Mar 2014, 00:23
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If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x? (1) z is prime (2) x is prime
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Originally posted by jimjohn on 14 Dec 2007, 12:31.
Last edited by Bunuel on 31 Mar 2014, 00:23, edited 2 times in total.
Added the OA.




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Re: If x, y, and z are integers greater than 1 and (3^27)(35^10)
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20 Mar 2014, 21:09
goodyear2013 wrote: If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x? (1) z is prime (2) x is prime OE: (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y) Break up the 35^10 and simplify the 9^14 (3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) Divide both sides by common terms 5^8, 7^10, 3^27 (5^2)(z) = (3)(x^y)
(1) SUFFICIENT: z must have a factor of 3 to balance the 3 on the right side of the equation. (1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3. Since z = 3, the left side of the equation is 75, so x^y = 25. The only integers greater than 1 that satisfy this equation are x = 5 and y = 2, so (1) is sufficient. Put differently, the expression x^y must provide the two fives that we have on the left side of the equation. The only way to get two fives if x and y are integers greater than 1 is if x = 5 and y = 2.
(2) SUFFICIENT: x must have a factor of 5 to balance out the 5^2 on the left side. Since (2) says that x is prime, x cannot have any other factors, so x = 5. Hi, can anyone explain how this works, please. Split everything into prime factors: \((3^{27})(35^{10})(z) = (5^8)(7^{10})(9^{14})(x^y)\) \((3^{27})(5^{10})(7^{10})*(z) = (3^{28})(5^8)(7^{10})(x^y)\) Now powers of prime factors on both sides of the equation should match since all variables are integers. If you have only \(3^{27}\) on left hand side, it cannot be equal to the right hand side which has \(3^{28}\). Prime factors cannot be created by multiplying other numbers together and hence you must have the same prime factors with the same powers on both sides of the equation. Stmnt 1: z is prime Note that you have \(3^{28}\) on Right hand side but only \(3^{27}\) on left hand side. This means z must have at least one 3. Since z is prime, z MUST be 3 only. You get \((3^{28})(5^{10})(7^{10}) = (3^{28})(5^8)(7^{10})(x^y)\) Now \(5^2\) is missing on the right hand side since we have \(5^{10}\) on left hand side but only \(5^8\) on right hand side. So \(x^y\) must be \(5^2\). x MUST be 5. Sufficient. Stmnt2: x is prime If x is prime, it must be 5 since \(5^2\) is missing on the right hand side. This would give us \(x^y = 5^2\). Sufficient. Answer (D)
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jimjohn wrote: oh sorry guys i didnt notice that the exponents didnt appear. plz note the edited question
In this case: D
5^2*z=3*x^y
1. z is prime and is 3. So, x=5 SUFF.
2. x is prime and is 5. So, x=5 SUFF.
but (327)*(3510)*(z) = (58)*(710)*(914)*(xy) is a toplevel problem



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Re: DS  primes
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14 Dec 2007, 18:20
jimjohn wrote: If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime
(3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y)
rewritten as
(3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y)
simplified to
25z=3(x^y) x^y is an integer, multiple of 25 so z is a multiple of 3
I z prime, so 3; x=5
II x is prime, x^y multiple of 25 so x can only be 5
D



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ok so i understand up until 25 * z = 3 * (x^y)
now how do we know that z has to be a multiple of 3. is it because we know that 3 is a factor of 25 * z, and since 3 is not a factor of 25 then it has to be a factor of z.
is there such a rule like that, that if x is a factor of a*b, then x has to be a factor of one of a or b.
thx



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If x, y, and z are integers greater than 1 and (3^27)(35^10)
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20 Mar 2014, 13:44
If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x? (1) z is prime (2) x is prime OE: (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y) Break up the 35^10 and simplify the 9^14 (3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) Divide both sides by common terms 5^8, 7^10, 3^27 (5^2)(z) = (3)(x^y)
(1) SUFFICIENT: z must have a factor of 3 to balance the 3 on the right side of the equation. (1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3. Since z = 3, the left side of the equation is 75, so x^y = 25. The only integers greater than 1 that satisfy this equation are x = 5 and y = 2, so (1) is sufficient. Put differently, the expression x^y must provide the two fives that we have on the left side of the equation. The only way to get two fives if x and y are integers greater than 1 is if x = 5 and y = 2.
(2) SUFFICIENT: x must have a factor of 5 to balance out the 5^2 on the left side. Since (2) says that x is prime, x cannot have any other factors, so x = 5. Hi, can anyone explain how this works, please.



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Re: If x, y, and z are integers greater than 1 and (3^27)(35^10)
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21 Mar 2014, 00:44
goodyear2013 wrote: If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x? (1) z is prime (2) x is prime OE: (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y) Break up the 35^10 and simplify the 9^14 (3^27)(5^10)(7^10)(z) = (5^8)(7^10)(3^28)(x^y) Divide both sides by common terms 5^8, 7^10, 3^27 (5^2)(z) = (3)(x^y)
(1) SUFFICIENT: z must have a factor of 3 to balance the 3 on the right side of the equation. (1) says that z is prime, so z cannot have another factor besides the 3. Therefore z = 3. Since z = 3, the left side of the equation is 75, so x^y = 25. The only integers greater than 1 that satisfy this equation are x = 5 and y = 2, so (1) is sufficient. Put differently, the expression x^y must provide the two fives that we have on the left side of the equation. The only way to get two fives if x and y are integers greater than 1 is if x = 5 and y = 2.
(2) SUFFICIENT: x must have a factor of 5 to balance out the 5^2 on the left side. Since (2) says that x is prime, x cannot have any other factors, so x = 5. Hi, can anyone explain how this works, please. Similar question to practice: ifxyandzareintegersgreaterthan1and90644.htmlHope it helps.
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walker wrote: jimjohn wrote: 1. z is prime and is 3. So, x=5 SUFF.
Aren't there 2 possible answers for #1? x=5, y=2 x=25, y=1 Thanks!



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Bunuel wrote: we are told that x, y, and z are integers greater than 1 Oh! Can't believe I missed that! It makes sense now . Thanks!



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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10
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01 Sep 2014, 16:11
Can anyone explain whether my approach is valid? 5^2*z = 3*x^y (x^y)/z = (5^2)/3 = (5^2a)/(3a) x^y = 5^2a z = 3a
(1) z is prime, so a = 1 and x^y = 25 => x = 5 S
(2) x is prime, so a = 1 and z = 3 S
D



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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10
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09 Oct 2015, 13:31
VeritasPrepKarishma wrote: Stmnt2: x is prime If x is prime, it must be 5 since \(5^2\) is missing on the right hand side. This would give us \(x^y = 5^2\). Sufficient.
Answer (D)
I just wanted to point out that \(x^y = 5^2\) is not necessarily true. In fact, if the question asked for the value of y, then statement 2 would have been insufficient\((3^{27})(35^{10})*(z) = (5^8)(7^{10})(9^{14})(x^y)\) can be rewritten as \(\frac{(5^2 *z)}{3}=x^y\)Written in this form, it is easy to notice that z must be a multiple of 3 since \(x^y\) is an integer. Since it is given that x is prime, the prime factorization of \(x^y\) will be x repeated y times, which means z should have at most one 3. Statement 2 doesn't require z to be prime, it could have a prime factorization of one 3 with any number of 5's and x must be 5, but y could take many integer values besides 2.For example: z=3*\(5^2\), x=5, y=4z=3*\(5^3\), x=5, y=5z=3*\(5^4\), x=5, y=6x must be 5, so statement 2 is sufficient.



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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10
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09 Oct 2015, 14:30
jimjohn wrote: If x, y, and z are integers greater than 1, and (3^27)(35^10)(z) = (5^8)(7^10)(9^14)(x^y), then what is the value of x?
(1) z is prime
(2) x is prime I found an almost identical question with the same question stem, but different statements. Here it is: ifxyandzareintegersgreaterthan1and90644.html



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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10
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07 Feb 2016, 10:19
There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay



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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10
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08 Apr 2016, 21:31
MrSobe17 wrote: There is one thing I don't understand about this problem and would appreciate any help.
When we simplify the equation and get 5^2 * (z) = 3 * (x^y), why do we even need any of the statements in the first place? Why isn't it directly clear that z is 3, x is 5 and y is 2? I really don't get it.
Thank you so much in advance.
Jay z can take any value in that case. Think of a case in which z = 12. \(5^2 * 3 * 2^2 = 3 * x^y\) Here x = 10
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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10
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