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

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Manager
Joined: 19 Aug 2007
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If x, y, and z are integers greater than 1, and (3^27)(35^10  [#permalink]

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Updated on: 31 Mar 2014, 00:23
5
36
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Difficulty:

95% (hard)

Question Stats:

45% (02:24) correct 55% (02:32) wrong based on 860 sessions

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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

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.
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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, 21:09
10
3
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.

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14 Dec 2007, 13:58
4
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
Manager
Joined: 01 Nov 2007
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14 Dec 2007, 18:20
2
1
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, 19:29
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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Joined: 21 Oct 2013
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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, 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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Joined: 02 Sep 2009
Posts: 52344
Re: If x, y, and z are integers greater than 1 and (3^27)(35^10)  [#permalink]

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21 Mar 2014, 00:44
1
1
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: if-x-y-and-z-are-integers-greater-than-1-and-90644.html

Hope it helps.
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Joined: 14 Mar 2014
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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!
Math Expert
Joined: 02 Sep 2009
Posts: 52344

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31 Mar 2014, 00:26
1
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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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]

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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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Joined: 01 Jan 2015
Posts: 63
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10  [#permalink]

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09 Oct 2015, 13:31
1
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.

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.
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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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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: if-x-y-and-z-are-integers-greater-than-1-and-90644.html
Manager
Joined: 12 Sep 2015
Posts: 80
Re: If x, y, and z are integers greater than 1, and (3^27)(35^10  [#permalink]

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07 Feb 2016, 10:19

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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Joined: 16 Oct 2010
Posts: 8803
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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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08 Apr 2016, 21:31
MrSobe17 wrote:

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  [#permalink]

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19 May 2018, 23:23
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