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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

(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

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.

(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.

(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.

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

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.

(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.

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

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09 Oct 2015, 07:37

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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, 13:31

1

This post received KUDOS

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.

Answer (D)

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.

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

There is one thing I don't understand about this problem and would appreciate any help.

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.

There is one thing I don't understand about this problem and would appreciate any help.

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.

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