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 and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48 B. 36 C. 24 D. 18 E. 12

Notice that we are told that \(x\) and \(y\) are positive integers.

\(5^x-5^y=2^{y-1}*5^{x-1}\);

\(5^x-2^{y-1}*5^{x-1}=5^y\);

\(5^x(1-\frac{2^y}{2}*\frac{1}{5})=5^y\);

\(5^x(10-2^y)=2*5^{y+1}\).

Now, since the right hand side is always positive then the left hand side must also be positive, hence \(10-2^y\) must be positive, which means that \(y\) can take only 3 values: 1, 2 and 3.

By trial and error we can find that only \(y=3\) gives integer value for \(x\):

Now, since the right hand side is always positive then the left hand side must also be positive, hence \(10-2^y\) must be positive, which means that \(y\) can take only 3 values: 1, 2 and 3.

By trial and error we can find that only \(y=3\) gives integer value for \(x\): \(5^x(10-2^3)=2*5^{3+1}\) --> \(2*5^x=2*5^4\) --> \(x=4\) --> \(xy=12\).

Answer: E.

Bunuel, I have a question, how did you know that you had to reorganize the equation in this way?

From \(5^x-5^y=2^{y-1}*5^{x-1}\) TO \(5^x-2^{y-1}*5^{x-1}=5^y\)

Now, since the right hand side is always positive then the left hand side must also be positive, hence \(10-2^y\) must be positive, which means that \(y\) can take only 3 values: 1, 2 and 3.

By trial and error we can find that only \(y=3\) gives integer value for \(x\): \(5^x(10-2^3)=2*5^{3+1}\) --> \(2*5^x=2*5^4\) --> \(x=4\) --> \(xy=12\).

Answer: E.

Bunuel, I have a question, how did you know that you had to reorganize the equation in this way?

From \(5^x-5^y=2^{y-1}*5^{x-1}\) TO \(5^x-2^{y-1}*5^{x-1}=5^y\)

Thanks!

Not directed at me, However you can re-arrange it another way.

We have \(5^x-5^y = 2^{y-1}*5^{x-1}\), Dividing on both sides by \(5^{x-1}\), we have \(5-5^{y+1-x} = 2^{y-1} \to 5 = 5^{y+1-x} + 2^{y-1}\). Now, as x and y are positive integers, the only value which \(2^{y-1}\) can take is 4.Thus, y-1 = 2, y = 3. Again, the value of \(5^{y+1-x}\) has to be 1, thus, y+1-x = 0 \(\to\) x = y+1 = 3+1 = 4. Thus, x*y = 4*3 = 12.

Thus, I think you could re-arrange in any order, as long as you get a tangible logic.
_________________

If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^(x-1), what is the value of xy?

A. 48 B. 36 C. 24 D. 18 E. 12

A little bit of observation can help you solve this question within a minute.

x and y are positive integers which means we will have clean numbers. On the right hand side, you have a 2 as a factor while it is not there on the left hand side. Can a 2 be generated on the left hand side by the subtraction? Here I am thinking that if we take 5^y common on the left hand side, I might be able to get a 2.

\(5^y (5^{x-y} - 1) = 2^{y-1}*5^{x-1}\) Now I want only 2s and 5s on the left hand side. If x-y is 1, then \((5^{x-y} - 1)\) becomes 4 which is 2^2. If instead x - y is 2 or more, I will get factors such as 3, 13 too. So let me try putting x - y = 1 to get \(5^y (2^2) = 2^{y-1}*5^{x-1}\)

This gives me y - 1 = 2 y = 3 x = 4 Check to see that the equations is satisfied with these values. Hence xy = 12

Answer (E)

Note that it is obvious that y is less than x and that is the reason we took \(5^y\) common. The reason it is obvious is that the right hand side is positive. So the left hand side must be positive too. This means \(5^x > 5^y\) which means x > y.
_________________

Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]

Show Tags

19 Feb 2014, 22:50

6

This post received KUDOS

Faster: First constrain the possible answers. We know \(x>y\) since if \(x=y\) then the left-hand side is 0 or if \(x<y\) then the LHS is negative... but the RHS is always positive. Now act: divide both sides by \(5^{x-1}\) to get \(5-5^{y-x+1} = 2^{y-1}\).

Since the new RHS is a power of two, the LHS must equal 1, 2, or 4. The only power of 5 that gets us one of those is \(5-5^0=5-1=4\). That means \(y=3\) and thus \(x=4\).

Oh, and BTW: The thread title is different than the equality in your post. (Title should have 5^{x-1}, not 5^x.)

Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]

Show Tags

19 Feb 2014, 23:01

@Karishma, thanks again.

@SizeTrader, appreciate the solution and that reasoning was excellent. Also, the reason the title says 5^x is because I reached the character limit for the title and it got cut off
_________________

Any and all kudos are greatly appreciated. Thank you.

Re: If x and y are positive integers and 5^x [#permalink]

Show Tags

20 Feb 2014, 15:02

If x and y are positive integers and (5^x)-(5^y)=(2^{y-1})*(5^{x-1}), what is the value of xy?

A. 48 B. 36 C. 24 D. 18 E. 12

E

Protocol:Simplify expression

1. need to find X and Y but cant isolate X and Y directly so start by separating the bases:

divide both sides by (5^{x-1}

left and right side simplifies to 5-5^{y-x+1} = 2^{y-1}

2. after simplifying, analyze.

Right side must be positive integer ( y is at least 1 ) thus left side must be positive too. ==> Left side is 5 minus an expression so answer must be at max 4 and at minimum 1. Right side must also be equal to a power of 2. Thus 4 is only possible answer for left side. Thus Y = 3.

Note: From original expression it is clear that X> Y. However by the time you get to simplifying the expression, the possible number of answers is so constrained that this information isn't critical to arriving at a faster answer.
_________________

Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]

Show Tags

24 Feb 2014, 23:57

SizeTrader wrote:

Faster: First constrain the possible answers. We know \(x>y\) since if \(x=y\) then the left-hand side is 0 or if \(x<y\) then the LHS is negative... but the RHS is always positive. Now act: divide both sides by \(5^{x-1}\) to get \(5-5^{y-x+1} = 2^{y-1}\).

Since the new RHS is a power of two, the LHS must equal 1, 2, or 4. The only power of 5 that gets us one of those is \(5-5^0=5-1=4\). That means \(y=3\) and thus \(x=4\).

Oh, and BTW: The thread title is different than the equality in your post. (Title should have 5^{x-1}, not 5^x.)

If x and y are positive integers and 5^x [#permalink]

Show Tags

19 Aug 2015, 07:30

Bunuel wrote:

BN1989 wrote:

If x and y are positive integers and (5^x)-(5^y)=(2^(y-1))*(5^(x-1)), what is the value of xy?

A. 48 B. 36 C. 24 D. 18 E. 12

Notice that we are told that \(x\) and \(y\) are positive integers.

\(5^x-5^y=2^{y-1}*5^{x-1}\);

\(5^x-2^{y-1}*5^{x-1}=5^y\);

\(5^x(1-\frac{2^y}{2}*\frac{1}{5})=5^y\);

\(5^x(10-2^y)=2*5^{y+1}\).

Now, since the right hand side is always positive then the left hand side must also be positive, hence \(10-2^y\) must be positive, which means that \(y\) can take only 3 values: 1, 2 and 3.

By trial and error we can find that only \(y=3\) gives integer value for \(x\):

\(5^x(10-2^3)=2*5^{3+1}\);

\(2*5^x=2*5^4\);

\(x=4\) --> \(xy=12\).

Answer: E.

i did another logic is it right?

\((5^x)(1-(2^y-1)/5)=5^y\)

then \((5^x-y-1)(5-(2^y-1))=1\)

which means this must be \(x-y-1= 0\) and \(5-(2^y-1) = 1\) means also \(2^y-1 = 4\) then \(y-1=2\) then y=3 replace y in old equation we get x =4 then finally xy =12

Re: If x and y are positive integers and 5^x [#permalink]

Show Tags

26 Aug 2016, 18:47

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 and y are positive integers and 5^x [#permalink]

Show Tags

21 Sep 2016, 02:31

Hi, The explanations are great. However, I took a lot of time figuring out how to simplify the equation and get to some solution and then started exploring otherway around which worked faster for me..

Difference between any 2 powers of 5 would always yields an even number.,i.e., a multiple of 2. However, difference of only consecutive powers of 5 yields a number that is only multiple of 2 and 5. Check->(25-5), (125-25), (625-125). Also check(625-25), (625-5), etc. Then, I simply had to check the options which had consecutive integers as factors. Only 12 worked out with 3 and 4 as factors.

This is not a foolproof solution but just another way of thinking incase you feel trapped in a question.

gmatclubot

Re: If x and y are positive integers and 5^x
[#permalink]
21 Sep 2016, 02:31

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...