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Manhattan GMAT CAT- The Powers That Be (2)

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Manhattan GMAT CAT- The Powers That Be (2)

by Misterbuff » Tue Sep 23, 2008 5:24 pm
Hi everyone, please see the problem below:
If x, y, and z are integers greater than 1, and (3^27)(5^ 10)(z) = (5^8)(9^14)(x^y), then what is the value of x?

(1) y is prime

(2) x is prime

According to Manhattan GMAT, it's B, but I think it's E.

I read the explanation, but I still believe it's E because I have a counterexample. If you have a good explanation that justifies the Manhattan GMAT answer, please enlighten me.
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Re: Manhattan GMAT CAT- The Powers That Be (2)

by Morgoth » Tue Sep 23, 2008 6:56 pm
Misterbuff wrote:Hi everyone, please see the problem below:
If x, y, and z are integers greater than 1, and (3^27)(5^ 10)(z) = (5^8)(9^14)(x^y), then what is the value of x?

(1) y is prime

(2) x is prime

According to Manhattan GMAT, it's B, but I think it's E.

I read the explanation, but I still believe it's E because I have a counterexample. If you have a good explanation that justifies the Manhattan GMAT answer, please enlighten me.
(3^27)(5^ 10)(z) = (5^8)(9^14)(x^y)
This can be simplified as
(3^27)(5^ 10)(z) = (5^8)(3^28)(x^y)

1) y is prime

well z has to be at least 3 and y can be 2,3, anything, we dont know anything about x. Insufficient.

2) x is prime

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

we know z has to be at least 3, x is prime, x = 5, y= 2

3^27 * 5^10 * 3 = 5^8 * 3^28 * 5^2

sufficient.

B is the answer.
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by Misterbuff » Tue Sep 23, 2008 7:11 pm
Wow, I wrote up my reasoning and then I just realized my error. I was thinking of the value of y instead of x. Thanks for the help!
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by 4meonly » Wed Sep 24, 2008 7:27 am
I used the same approach as Morgoth but I'll try add more reasoning about A

(3^27)(5^ 10)(z) = (5^8 )(9^14)(x^y)
This can be simplified as
(3^27)(5^ 10)(z) = (5^8 )(3^28 )(x^y)
and finally, after dividing bot parts by
(3^27)(5^8 )
we'll get
(5^2)(z) = (3)(x^y)
z should ne equal to 3 (because of the power and because each part of the equation should contain same prime factors with same powers)
so, finally, we get
(5^2) = (x^y)
what is the value of x?
if x=5, y=2
if x=25, y=1


(1) y is prime
if x=5, y=2
if x=25, y=1 (1 is prime? this is debatable question)
INSUFF


(2) x is prime
means that
y=2, x=5
SUFF

B
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by Mahalo » Fri Oct 03, 2008 12:04 pm
I am arriving at (5^2)/3 = (x^y)/z;
This leaves Z=3, x=5 and y=2; Since all the numbers are greater than 1, x=25 and y=1 option is eliminated..

So the answer is D? can someone explain , why the answer is not D?
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by conomav » Fri Oct 03, 2008 12:36 pm
for eliminating A here is explanation
simplifying we get

5^2 *z = 3 * x^y

1) y is prime

say y=2
now x can be any multiple of 5 as smallest prime is 2
picking some values
for y = 2
x = 5 z= 3
x = 10 z=12

so x can be changing & hence insufficient

ii) x is prime
only options is 5 as a 5 as a prime factor is needed.

B sufficient
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by Mahalo » Fri Oct 03, 2008 12:43 pm
Thanks conomav
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by raju232007 » Fri Oct 03, 2008 12:46 pm
3^27*5^10*z=5^8*9^14*x^y
This can be simplified as
25z=3x^y

statement 1 says that y is prime

assume y=2 then x=5 and z=3
y=1 then x=25 and z=3(This case should not be considered as 2melony had pointed out)
In this case we know that y is prime but we cannot determine the value of x&z..For instance if y=3 ,x=0 & z=0 the condtion still holds true..so x=0,5 are the values of x...Insufficient

Statement 2 says that x is prime

In this case x must be 5,which is prime to balance the prime factors o both the sides...

And moreover the question being asked is to determine the value of x...So we dont have to bother about y or z...So the only possible solution for x is 5...Statement 2 is sufficient...

Hope this explanation helps.. Let me know if u still got any doubts..
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by dally_gmat » Fri Oct 03, 2008 4:13 pm
Can some one help me explain why z cannot be equal to 3..??
if z = 3, x = 5, y =2, then eq is satisfied ...I agree that x can be 25 and y can be one...

Please help..
Thanks

5^2)(z) = (3)(x^y)
z should ne equal to 3 (because of the power and because each part of the equation should contain same prime factors with same powers)

raju232007 wrote:3^27*5^10*z=5^8*9^14*x^y
y=1 then x=25 and z=3(This case should not be considered as 2melony had pointed out)
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by 4meonly » Sun Oct 05, 2008 4:18 am
dally_gmat wrote:Can some one help me explain why z cannot be equal to 3..??
Z IS equal to 3
(5^2)(z) = (3)(x^y)
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by nitin86 » Sun Oct 05, 2008 7:27 pm
4meonly wrote:I used the same approach as Morgoth but I'll try add more reasoning about A

(3^27)(5^ 10)(z) = (5^8 )(9^14)(x^y)
This can be simplified as
(3^27)(5^ 10)(z) = (5^8 )(3^28 )(x^y)
and finally, after dividing bot parts by
(3^27)(5^8 )
we'll get
(5^2)(z) = (3)(x^y)
z should ne equal to 3 (because of the power and because each part of the equation should contain same prime factors with same powers)
so, finally, we get
(5^2) = (x^y)
what is the value of x?
if x=5, y=2
if x=25, y=1


(1) y is prime
if x=5, y=2
if x=25, y=1 (1 is prime? this is debatable question)
INSUFF


(2) x is prime
means that
y=2, x=5
SUFF

B
Question stem says " x, y, and z are integers greater than 1..." so . we can't take y = 1
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by deepaksuresh » Mon Oct 06, 2008 2:48 pm
Why is the answer not D?y cannot be 1 for case 1 as 1 is not a prime number.
The only values possible are z=3 and x=5
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by 4meonly » Sun Jan 25, 2009 4:37 am
I made small corrections that prove that answer B is correct

3^27)(5^ 10)(z) = (5^8 )(9^14)(x^y)
This can be simplified as
(3^27)(5^ 10)(z) = (5^8 )(3^28 )(x^y)
and finally, after dividing bot parts by
(3^27)(5^8 )
we'll get
(5^2)(z) = (3)(x^y)
z is a multiply of 3 - it can be 3, 15, 75 etc. For example:

(5^2)(3) = (3)(x^y) => 5^2 = x^y => x=5, y=2
(5^2)(3*5) = (3)(5)(x^y) => 5^3 = x^y => x=5, y=3
(5^2)(3*5*5) = (3)(x^y) => 5^4 = x^y => x=5, y=4 OR
(5^2)(3*5*5) = (3)(x^y) => 5^4 = x^y => x=25, y=2
(5^2)(3*5^5) = (3)(x^y) => 5^7 = x^y => x=5, y=7

(1) y is prime
(5^2)(3) = (3)(x^y) => 5^2 = x^y => x=5, y=2
(5^2)(3*5) = (3)(5)(x^y) => 5^3 = x^y => x=5, y=3
(5^2)(3*5^5) = (3)(x^y) => 5^7 = x^y => x=5, y=7
INSUFF


(2) x is prime
(5^2)(3) = (3)(x^y) => 5^2 = x^y => x=5, y=2
the only value for x is 5 (because it is prime)
SUFF

B
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