## Primes/Exponents: If x, y, and z are integers greater than 1

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### Primes/Exponents: If x, y, and z are integers greater than 1

by II » Tue Feb 12, 2008 3:16 pm
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
Last edited by II on Tue Apr 29, 2008 7:27 am, edited 3 times in total.

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by its_me07 » Wed Feb 13, 2008 2:06 am
Could u please pos the question again.
whats the power of 5?

i think that OA shud b E as z can be any prime 2,3,5 ,7 n even if we find the value of z and x..we still don know he value of y.

whats the OA?

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by sankruth » Wed Feb 13, 2008 8:43 am
Im assuming it is 5^8

Simplifying the equation, you get

(25/3).z = x^y

Since x, y, z are integers greater than 1, x^y is also an integer.

So (25/3).z is an integer

St 1
z is prime

Only z = 3 satisfies the condition, (25/3).z is an integer

which gives x^y = 25 and x=5

SUFF

St 2
x is prime.

z must be a multiple of 3 for (25/3).z to be an integer.

x must be a multiple of 5.

Since x is prime and greater than 1, x can only be equal to 5

SUFF

So D

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by its_me07 » Wed Feb 13, 2008 9:07 am
plz post the OA

sankruth: we know 25=5^2 so if x^y=25 can we deduce the value of x ourselves as there is no inf given in st 1.

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by sankruth » Wed Feb 13, 2008 10:57 am
its_me07 wrote:plz post the OA

sankruth: we know 25=5^2 so if x^y=25 can we deduce the value of x ourselves as there is no inf given in st 1.
If x^y = 25 and x, y are integers there are two possibilities 25^1 and 5^2

But the question stem mentions that x, y and z are greater than 1, so it must be 5^2

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by II » Wed Feb 13, 2008 11:05 am
Thanks for your input guys ... but unfortunately I dont have the OA to this. It was sent to me by another friend of mine.

I am hoping Stuart Kovinsky (our resident number properties expert !) will provide some input to this thread.

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### Re: And another number properties ... dont you just love em

by Stuart@KaplanGMAT » Wed Feb 13, 2008 12:59 pm
II 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
The solutions posted are good, but sure, I'll throw my 2 cents in.

As always, when faced with a complicated stem we want to simplify it. Here, we see different very big numbers on both sides, so let's reduce to primes:

(3^27)(5^10)(7^10)z = (5^8)(7^10)(3^28)(x^y)

[35^10 = (5*7)^10 = 5^10 * 7^10]
[9^14 = (3^2)^14 = 3^(2*14) = 3^28]

Since we're solving for x, let's move all the numbers to the 'z' side. After reducing the exponents, we get:

(5^2)(z)/(3) = x^y

(25/3)z = x^y

As pointed out, we know that x, y and z are all integers greater than 1; therefore, x^y is an integer. Since the right side is an integer, the left side must also be an integer. For (25/3)z to be an integer, z MUST be a multiple of 3.

(1) z is prime.

Well, if z is a multiple of 3 AND z is prime, we know that z=3.

We now know that x^y = 25*3/3 = 25

If x and y are integers, there are two solutions:

a) x = 25 and y = 1; and
b) x = 5 and y = 2.

Since we know that y is greater than 1, x MUST be equal to 5. Sufficient!

(2) x is prime.

We could have written the original equation as:

z = (3/25)x^y

Since z is an integer, the right side must also be an integer. For the right side to be an integer, x^y has to be a multiple of 25.

For x^y to be a multiple of 25, x MUST be a multiple of 5 (since x and y are integers). If x is a multiple of 5 AND prime, x MUST be equal to 5. Sufficient!

Each statement is sufficient on its own: choose (d).

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BTG100 for $100 off a full course Master | Next Rank: 500 Posts Posts: 400 Joined: 10 Dec 2007 Location: London, UK Thanked: 19 times GMAT Score:680 by II » Thu Feb 14, 2008 1:53 am Thanks Stuart. Excellent clear explanation. Master | Next Rank: 500 Posts Posts: 400 Joined: 10 Dec 2007 Location: London, UK Thanked: 19 times GMAT Score:680 by II » Thu Feb 14, 2008 4:07 am Hi Stuart ... how would you classify this question ? Would you say it is easy (<550), medium (550-700), hard (700+) ? Thanks. ### GMAT/MBA Expert GMAT Instructor Posts: 3225 Joined: 08 Jan 2008 Location: Toronto Thanked: 1710 times Followed by:613 members GMAT Score:800 by Stuart@KaplanGMAT » Thu Feb 14, 2008 10:11 am II wrote:Hi Stuart ... how would you classify this question ? Would you say it is easy (<550), medium (550-700), hard (700+) ? Thanks. I'd estimate that this would be a mid-600s level question. Stuart Kovinsky | Kaplan GMAT Faculty | Toronto Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount BTG100 for$100 off a full course

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by II » Sun Feb 17, 2008 2:10 pm
Stuart Kovinsky wrote:
II wrote:Hi Stuart ... how would you classify this question ?
Would you say it is easy (<550), medium (550-700), hard (700+) ?

Thanks.
I'd estimate that this would be a mid-600s level question.
According to ManhattanGMAT, this is a 700+ level question.

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### Re: Primes/Exponents: If x, y, and z are integers greater th

by JasonReynolds » Sat May 23, 2009 4:57 pm
II 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
So the actual question that i got on Manhattan GMAT said that y was prime for statement (1), instead of z being prime. But it doesnt change a whole lot.

My main issue is, if you look at statement (2), how do you know that you cant have x = 5, y = 3, z = 15?

When you reduce the equations you have: 5^2 * z = 3x^y.
(5^2)(15) = 3(5^3)
(5^3)(3) = 3(5^3).

If I'm not mistaken, this is an alternate solution, that doesnt violate the information provided by Statement 2. Thus, statement 2 would not be sufficient?

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### Re: Primes/Exponents: If x, y, and z are integers greater th

by Stuart@KaplanGMAT » Mon May 25, 2009 10:07 am
JasonReynolds wrote:
II 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
So the actual question that i got on Manhattan GMAT said that y was prime for statement (1), instead of z being prime. But it doesnt change a whole lot.

My main issue is, if you look at statement (2), how do you know that you cant have x = 5, y = 3, z = 15?

When you reduce the equations you have: 5^2 * z = 3x^y.
(5^2)(15) = 3(5^3)
(5^3)(3) = 3(5^3).

If I'm not mistaken, this is an alternate solution, that doesnt violate the information provided by Statement 2. Thus, statement 2 would not be sufficient?

It's very important to identify exactly what the question is asking.

Here, the question is "what is the value of x?"

As you've pointed out, statement (2) gives us many possible values for y and z. However, we always get the same value for x (which has to be both prime and a multiple of 5), so the statement is sufficient to answer the question.

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by navalpike » Tue Jul 07, 2009 12:51 pm
When I initially read the question, I thought that one could solve the problem without reading any of the choices. I have read Ian mention the “Fundamental theorem of arithmetic” which (roughly) mentions that once the numbers on each side are reduced to their primes, their exponents must be equal on both sides. (please correct me if I have this wrong).

So, once we get –

5^2 (z) = x^y (3)

on each side, and we already know that x, y, and z are greater than 1, then according to the theorem, doesn’t everything has to match? Since 3 and z on each side are the only ones with 1 as their exponent, don’t they already match? (meaning z has to 3) And since 5^2 and x^y are the only ones on each side with an exponent higher than one, (y is greater than 1), don’t they already match? Why is it necessary for us to know that they are prime?

I am sure my thinking is flawed somehow, so I am grateful for you clearing it up.

Thanks,

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by adityanarula » Tue Jul 07, 2009 1:20 pm
Not Necessarily

As shown above z= (3/25)X^Y

If X is not prime, then X^Y could be any multiple of 25. For example, if X=10 & Y =2, then Z=(3/25)*100 = 12

Therefore we have to know if X is prime.