Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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]

Show Tags

09 Oct 2015, 08:37

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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

Show Tags

09 Oct 2015, 14:31

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]

Show Tags

07 Feb 2016, 11:19

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.

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

Show Tags

08 Apr 2016, 22:31

Expert's post

MrSobe17 wrote:

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.

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...