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) y is prime
(2) x is prime
Integer
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- theCodeToGMAT
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To find --> xvinay1983 wrote:If x, y, and z are integers greater than 1, and (327)(510)(z) = (58)(914)(xy), then what is the value of x?
(1) y is prime
(2) x is prime
327 * 510 * z / 58 * 914 * y = x
327 = 3 * 109
510 = 2 * 5 * 3 * 17
58 = 2 * 29
914 = 2 * 457
(3 * 109) * (5 * 3 * 17) * Z
________________________ = X
2 * 29 * 457 * y
Statement 1:
y is prime
X can have any value
INSUFFICIENT
Statement 2:
X is prime
y can have any value .. subsequently.. X can have any value out of 109,3,5,17
INSUFFICIENT
Combining...
y = 3 (Out of 109, 5, 3, 17)
Z = 457 * 29 * 2
x = NOT PRIME
I got stuck here...... What is the OA????
R A H U L
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I believe that the problem should read as follows:
(3²�)(7¹�5¹�)(z) = (5�)(7¹�)(3²)¹�(x^y)
(3²�)(7¹�5¹�)(z) = (5�)(7¹�)(3²�)(x^y)
(5²)(z) = (3)(x^y)
z = (3) * (x^y)/5².
Since z is an INTEGER, the resulting equation implies that z is a multiple of 3 and that x^y is a multiple of 5².
Statement 1: z is prime
Since z is prime and a multiple of 3, z=3.
Thus, (x^y)/5² = 1, implying that x=5 and y=2.
SUFFICIENT.
Statement 2: x is prime
Since x^y is a multiple of 5² and x is prime, x=5 and y≥2.
SUFFICIENT.
The correct answer is D.
(3²�)(35¹�)(z) = (5�)(7¹�)(9¹�)(x^y)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
(3²�)(7¹�5¹�)(z) = (5�)(7¹�)(3²)¹�(x^y)
(3²�)(7¹�5¹�)(z) = (5�)(7¹�)(3²�)(x^y)
(5²)(z) = (3)(x^y)
z = (3) * (x^y)/5².
Since z is an INTEGER, the resulting equation implies that z is a multiple of 3 and that x^y is a multiple of 5².
Statement 1: z is prime
Since z is prime and a multiple of 3, z=3.
Thus, (x^y)/5² = 1, implying that x=5 and y=2.
SUFFICIENT.
Statement 2: x is prime
Since x^y is a multiple of 5² and x is prime, x=5 and y≥2.
SUFFICIENT.
The correct answer is D.
Last edited by GMATGuruNY on Tue Oct 08, 2013 4:54 am, edited 1 time in total.
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- vinay1983
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Sorry. I am a little out of ordertheCodeToGMAT wrote:To find --> xvinay1983 wrote:If x, y, and z are integers greater than 1, and (327)(510)(z) = (58)(914)(xy), then what is the value of x?
(1) y is prime
(2) x is prime
327 * 510 * z / 58 * 914 * y = x
327 = 3 * 109
510 = 2 * 5 * 3 * 17
58 = 2 * 29
914 = 2 * 457
(3 * 109) * (5 * 3 * 17) * Z
________________________ = X
2 * 29 * 457 * y
Statement 1:
y is prime
X can have any value
INSUFFICIENT
Statement 2:
X is prime
y can have any value .. subsequently.. X can have any value out of 109,3,5,17
INSUFFICIENT
Combining...
y = 3 (Out of 109, 5, 3, 17)
Z = 457 * 29 * 2
x = NOT PRIME
I got stuck here...... What is the OA????
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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A student pointed me to Mitch's solution on this thread, in particular the part about statement 2.
Mitch --
However, I was wondering how you arrived at y = 2. For instance, why can't y = 3, in which case z = 15? Or y = 4, in which case z = 45? Etc. After all, we are no longer constrained by the requirement that z be prime.
Mitch --
It's true that x has to be 5. (Since we only care about x, this is all that matters for the problem at hand.)GMATGuruNY wrote:Statement 2: x is prime
Since x^y is a multiple of 5² and x is prime, x=5 and y=2.
SUFFICIENT.
However, I was wondering how you arrived at y = 2. For instance, why can't y = 3, in which case z = 15? Or y = 4, in which case z = 45? Etc. After all, we are no longer constrained by the requirement that z be prime.
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Pesky typos! The solution should have indicated that y≥2 (since x^y must be a multiple of 5² and x must be prime). Edited my reply.lunarpower wrote:However, I was wondering how you arrived at y = 2. For instance, why can't y = 3, in which case z = 15? Or y = 4, in which case z = 45? Etc. After all, we are no longer constrained by the requirement that z be prime.
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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I unlock the best way for YOU to solve problems.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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