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.

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:

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.

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. _________________

Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]
19 Feb 2014, 21:25

4

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

MrWallSt 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

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]
19 Feb 2014, 21:50

4

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]
19 Feb 2014, 22:01

@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]
20 Feb 2014, 14:02

Expert's post

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]
24 Feb 2014, 22:57

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.)

Nicely explained. _________________

Kindly press "+1 Kudos" to appreciate

gmatclubot

Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x
[#permalink]
24 Feb 2014, 22:57