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# If x and y are positive integers and 5^x

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Manager
Joined: 12 Oct 2011
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If x and y are positive integers and 5^x [#permalink]

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06 Apr 2012, 09:42
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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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 44401
If x and y are positive integers and 5^x [#permalink]

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06 Apr 2012, 14:01
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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 $$x$$:

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

$$2*5^x=2*5^4$$;

$$x=4$$ --> $$xy=12$$.

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

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24 Sep 2013, 15:38
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 $$x$$: $$5^x(10-2^3)=2*5^{3+1}$$ --> $$2*5^x=2*5^4$$ --> $$x=4$$ --> $$xy=12$$.

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!
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 624
Re: If x and y are positive integers and 5^x [#permalink]

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24 Sep 2013, 23:04
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danzig wrote:
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 $$x$$: $$5^x(10-2^3)=2*5^{3+1}$$ --> $$2*5^x=2*5^4$$ --> $$x=4$$ --> $$xy=12$$.

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

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19 Feb 2014, 22:25
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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

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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 19 Feb 2014 Posts: 18 Concentration: Finance GRE 1: 1580 Q800 V780 WE: Securities Sales and Trading (Investment Banking) 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 7 This post received KUDOS 1 This post was BOOKMARKED 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.) Intern Joined: 21 Nov 2013 Posts: 12 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. Senior Manager Joined: 13 Jun 2013 Posts: 278 Re: If x and y are positive integers and 5^x [#permalink] ### Show Tags 20 Feb 2014, 04:53 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 by dividing both sides of the equation with 5^x we have 1- 5^(y-x) = 2^(y-1) * 5^(x-1-x) 1- 5^(y-x) = 2^(y-1) * 5^(-1) 5= 2^(y-1) + 5^(y-x+1) now minimum value of 2^(y-1) =1, hence 2^(y-1) must be equal to 4 and 5^(y-x+1) must be equal to 1 for the R.H.S to become equal to L.H.S. 2^(y-1) = 4 for y=3 and 5^(y-x+1) =1 for x=4 (as y=3) hence product of xy = 12 GMAT Tutor Joined: 20 Jul 2012 Posts: 25 GMAT 1: 780 Q50 V50 Re: If x and y are positive integers and 5^x [#permalink] ### Show Tags 20 Feb 2014, 15:02 1 This post received KUDOS 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. _________________ WWW.CLEARMOUNTAINPREP.COM Intern Joined: 02 Mar 2015 Posts: 32 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 Intern Joined: 21 Nov 2014 Posts: 40 Location: India Concentration: General Management, Strategy GMAT 1: 750 Q51 V40 WE: Operations (Energy and Utilities) Re: If x and y are positive integers and 5^x [#permalink] ### Show Tags 19 Aug 2015, 08:04 (5^x)-(5^y)=(2^(y-1))*(5^(x-1)) (5^x)(1-5^y-x) = (2^(y-1))*(5^(x-1)) 5(1-5^y-x) = (2^(y-1)) 10(1-5^y-x) = (2^(y)) We note that (2^(y)) is always positive. which translates to (1-5^y-x) >0 or y-x<0 So, 10(5^y-x)(5^(x-y)-1) = (2^(y)) 5^(y-x+1)*2*(5^(x-y)-1) =(2^(y)) Now, RHS is 5^0, which means y-x+1 = 0, x-y =1 inputting values in above, 5^(0)*2*(5^(1)-1) =(2^(y)) 2*4 = (2^(y)) implies y =3 x= 4 xy = 12 Intern Joined: 18 Jul 2016 Posts: 9 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. SVP Joined: 12 Sep 2015 Posts: 2156 Location: Canada Re: If x and y are positive integers and 5^x [#permalink] ### Show Tags 01 Aug 2017, 09:57 Expert's post Top Contributor 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 Given: $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$ Divide both sides by $$5^{x-1}$$ to get: $$5^1 - 5^{y-x+1} = 2^{y-1}$$ Simplify: $$5 - 5^{y-x+1} = 2^{y-1}$$ OBSERVE: Notice that the right side, $$2^{y-1}$$, is POSITIVE for all values of y Since y is a positive integer, $$2^{y-1}$$ can equal 1, 2, 4, 8, 16 etc (powers of 2) So, the left side, $$5 - 5^{y-x+1}$$, must be equal 1, 2, 4, 8, 16 etc (powers of 2). Since $$5^{y-x+1}$$ is always positive, we can see that $$5 - 5^{y-x+1}$$ cannot be greater than 5 So, the only possible values of $$5 - 5^{y-x+1}$$ are 1, 2 or 4 In other words, it must be the case that: case a) $$5 - 5^{y-x+1} = 2^{y-1} = 1$$ case b) $$5 - 5^{y-x+1} = 2^{y-1} = 2$$ case c) $$5 - 5^{y-x+1} = 2^{y-1} = 4$$ Let's test all 3 options. case a) $$5 - 5^{y-x+1} = 2^{y-1} = 1$$ This means y = 1 (so that the right side evaluates to 1) The left side, $$5 - 5^{y-x+1} = 1$$, when $$5^{y-x+1} = 4$$. Since x and y are positive integers, it's IMPOSSIBLE for $$5^{y-x+1}$$ to equal 4 So, we can eliminate case a case b) $$5 - 5^{y-x+1} = 2^{y-1} = 2$$ This means y = 2 (so that the right side evaluates to 2) The left side, $$5 - 5^{y-x+1} = 2$$, when $$5^{y-x+1} = 3$$. Since x and y are positive integers, it's IMPOSSIBLE for $$5^{y-x+1}$$ to equal 3 So, we can eliminate case b case c) $$5 - 5^{y-x+1} = 2^{y-1} = 4$$ This means y = 3 (so that the right side evaluates to 4) The left side, $$5 - 5^{y-x+1} = 4$$, when $$5^{y-x+1} = 1$$. If $$5^{y-x+1} = 1$$, then $$y-x+1 = 0$$ In this case, y = 3 So, we can write: 3 - x + 1 = 0, which mean x = 4 So, the only possible solution is y = 3 and x = 4, which means xy = (4)(3) = 12 Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com Director Joined: 13 Mar 2017 Posts: 573 Location: India Concentration: General Management, Entrepreneurship GPA: 3.8 WE: Engineering (Energy and Utilities) Re: If x and y are positive integers and 5^x [#permalink] ### Show Tags 07 Aug 2017, 22:37 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 $$(5^x)-(5^y)=(2^{y-1})*(5^{x-1})$$ =$$(5^{x-1})(5-5^{y-x+1})=(2^{y-1})*(5^{x-1})$$ 5^{x-1} =/=0. So, $$(5-5^{y-x+1})=(2^{y-1})$$ Since $$(2^{y-1})$$ must be +ve. Also Y is +ve so, y-1>0 and hence $$(2^{y-1})$$ will be an integer only. Hence y-x+1 can be 0 only. y-x+1 = 0 -> y-x = -1 also at y-x+1 = 0 $$5-1 = (2^{y-1})$$ $$4 = (2^{y-1})$$ y -1 = 2 y = 3 x= y+1 = 4 xy = 12 Answer E _________________ CAT 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95 UPSC Aspirants : Get my app UPSC Important News Reader from Play store. MBA Social Network : WebMaggu Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish". EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11309 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: If x and y are positive integers and 5^x [#permalink] ### Show Tags 30 Jan 2018, 14:30 Hi All, GMAT Quant questions can almost always be solved in a variety of ways, so if you find yourself not able to solve by doing complex-looking math, then you should look for other ways to get to the answer. Think about what's in the question; think about how the rules of math "work." This question is LOADED with Number Property clues - when combined with a bit of "brute force", you can answer this question by doing a lot of little steps. Here are the Number Properties (and the deductions you can make as you work through them): 1) We're told that X and Y are POSITIVE INTEGERS, which is a great "restriction." 2) We can calculate powers of 5 and powers of 2 rather easily: 5^0 = 1 5^1 = 5 5^2 = 25 5^3 = 125 5^4 = 625 Etc. 2^0 = 1 2^1 = 2 2^2 = 4 Etc. 3) The answer choices are ALL multiples of 3. Since we're asked for the value of XY, either X or Y (or both) MUST be a multiple of 3. 4) *The left side of the equation is a positive number MINUS a positive number. *The right side is the PRODUCT of two positive numbers, which is POSITIVE. *This means that 5^X > 5^Y, so X > Y. 5) *Notice how we have 5^X (on the left side) and 5^(X-1) on the right side; these are consecutive powers of 5, so the first is 5 TIMES bigger than the second. *On the left, we're subtracting a number from 5^X. *On the right, we're multiplying 2^(Y-1) times 5^(X-1). *2^(Y-1) has to equal 1, 2 or 4, since if it were any bigger, then multiplying by that value would make the right side of the equation TOO BIG (you'd have a product that was bigger than 5^X). *By extension, Y MUST equal 1, 2 or 3. It CANNOT be anything else. 6) Remember that at least one of the variables had to be a multiple of 3. What if Y = 3? Let's see what happens…. 5^X - 5^3 = 2^2(5^(X-1)) 5^X - 125 = 4(5^(X-1)) Remember that X > Y, so what if X = 4?….. 5^4 - 125 = 500 4(5^3) = 500 The values MATCH. This means Y = 3 and X = 4. XY = 12 Final Answer: [Reveal] Spoiler: E GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Re: If x and y are positive integers and 5^x [#permalink]

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31 Jan 2018, 02:10
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

Good Question.. I solved it this way..

Let $$x-1 = a$$, $$y-1 = b$$.. So the equation becomes,

$$5^a - 5^b$$ = $$\frac{5^a*2^b}{5}..$$
Now, we have to find the value of $$xy$$ i.e. $$(a+1)(b+1)$$
By looking at the options and the above equation only E satisfies.
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Re: If x and y are positive integers and 5^x [#permalink]

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02 Feb 2018, 12:09
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

We can simplify the given equation by dividing both sides by 5^(x - 1):

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

5 - 5^(y - x + 1) = 2^(y - 1)

It’s not easy to solve an equation with two variables by algebraic means; however, since both variables are positive integers, we can try numbers for one of the variables and solve for the other.

If we let y = 1, then the right hand side (RHS) = 2^0 = 1 and thus the left hand side (LHS) is 5^(1 - x + 1) = 4. However, a power of 5 can’t be equal to 4 when the exponent is an integer.

Now, let’s let y = 2; then the RHS = 2^1 = 2 and thus, for the LHS, 5^(2 - x + 1) = 3. However, a power of 5 can’t be equal to 3 when the exponent is an integer.

Finally, let’s let y = 3; then the RHS = 2^2 = 4 and thus, for the LHS, 5^(3 - x + 1) = 1. We see that a power of 5 can be equal to 1 when the exponent is 0. Thus:

3 - x + 1 = 0

x = 4

We see that x = 4 and y = 3 and thus xy = 12.

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Re: If x and y are positive integers and 5^x   [#permalink] 02 Feb 2018, 12:09
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