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# If x, y, and z are integers greater than 1, and (3^27)(35^10

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Manager
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If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]

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14 Dec 2007, 13:31
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Mar 2014, 01:23, edited 2 times in total.
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14 Dec 2007, 14:58
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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 top-level problem
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14 Dec 2007, 19:20
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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)
re-written 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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14 Dec 2007, 20:29
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) [#permalink]

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20 Mar 2014, 14:44
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:
[Reveal] Spoiler:
(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) [#permalink]

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20 Mar 2014, 22:09
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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:
[Reveal] Spoiler:
(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.

_________________

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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 32557 Followers: 5641 Kudos [?]: 68413 [1] , given: 9805 Re: If x, y, and z are integers greater than 1 and (3^27)(35^10) [#permalink] ### Show Tags 21 Mar 2014, 01:44 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED 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: [Reveal] Spoiler: (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: if-x-y-and-z-are-integers-greater-than-1-and-90644.html Hope it helps. _________________ Intern Joined: 14 Mar 2014 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: [#permalink] ### Show Tags 30 Mar 2014, 12:23 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! Math Expert Joined: 02 Sep 2009 Posts: 32557 Followers: 5641 Kudos [?]: 68413 [0], given: 9805 Re: Re: [#permalink] ### Show Tags 31 Mar 2014, 01:26 Expert's post karimtajdin wrote: 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! Notice that we are told that x, y, and z are integers greater than 1, hence x=25 and y=1 is not possible. Check here for a complete solution: if-x-y-and-z-are-integers-greater-than-1-and-57122.html#p1346892 Similar question to practice: if-x-y-and-z-are-integers-greater-than-1-and-90644.html Hope it helps. _________________ Intern Joined: 14 Mar 2014 Posts: 2 Followers: 0 Kudos [?]: 0 [0], given: 1 Re: Re: [#permalink] ### Show Tags 31 Mar 2014, 06:02 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! Senior Manager Joined: 10 Mar 2013 Posts: 290 GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Followers: 3 Kudos [?]: 71 [0], given: 2404 Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink] ### Show Tags 01 Sep 2014, 17:11 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 GMAT Club Legend Joined: 09 Sep 2013 Posts: 9254 Followers: 455 Kudos [?]: 115 [0], given: 0 Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink] ### Show Tags 09 Oct 2015, 08:37 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Manager Joined: 01 Jan 2015 Posts: 62 Followers: 0 Kudos [?]: 35 [0], given: 14 Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink] ### Show Tags 09 Oct 2015, 14:31 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=4 z=3*$$5^3$$, x=5, y=5 z=3*$$5^4$$, x=5, y=6 x must be 5, so statement 2 is sufficient. Manager Joined: 01 Jan 2015 Posts: 62 Followers: 0 Kudos [?]: 35 [0], given: 14 Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink] ### Show Tags 09 Oct 2015, 15:30 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: if-x-y-and-z-are-integers-greater-than-1-and-90644.html Manager Joined: 12 Sep 2015 Posts: 112 Followers: 0 Kudos [?]: 4 [0], given: 24 Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink] ### Show Tags 07 Feb 2016, 11:19 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 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6477 Location: Pune, India Followers: 1758 Kudos [?]: 10481 [0], given: 206 Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink] ### Show Tags 08 Apr 2016, 22:31 Expert's post 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 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10   [#permalink] 08 Apr 2016, 22:31
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