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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]  14 Dec 2007, 12:31
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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, 00:23, edited 2 times in total.
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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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Re: DS - primes [#permalink]  14 Dec 2007, 18:20
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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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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]  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:
[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]  20 Mar 2014, 21:09
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Expert's post
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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Re: If x, y, and z are integers greater than 1 and (3^27)(35^10) [#permalink]  21 Mar 2014, 00:44
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Expert's post
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.
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Re: [#permalink]  30 Mar 2014, 11:23
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!
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Re: Re: [#permalink]  31 Mar 2014, 00:26
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.
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Re: Re: [#permalink]  31 Mar 2014, 05:02
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!
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Re: If x, y, and z are integers greater than 1, and (3^27)(35^10 [#permalink]  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
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10   [#permalink] 01 Sep 2014, 16:11
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