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If x and y are positive integers and 5^x

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If x and y are positive integers and 5^x [#permalink] New post 06 Apr 2012, 08:42
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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
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Re: If x and y are positive integers and 5^x [#permalink] New post 06 Apr 2012, 13:01
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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.

Answer: E.
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Re: If x and y are positive integers and 5^x [#permalink] New post 24 Sep 2013, 14:38
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.

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!
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Re: If x and y are positive integers and 5^x [#permalink] New post 24 Sep 2013, 22:04
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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.

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.
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If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink] New post 19 Feb 2014, 20:48
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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
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink] New post 19 Feb 2014, 21:25
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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

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.
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink] New post 19 Feb 2014, 21:50
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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.)
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink] New post 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
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Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink] New post 19 Feb 2014, 23:43
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Re: If x and y are positive integers and 5^x [#permalink] New post 20 Feb 2014, 03:53
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
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Re: If x and y are positive integers and 5^x [#permalink] New post 20 Feb 2014, 09:16
"Difficulty: 600-700 Level" seriously?! =/
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Re: If x and y are positive integers and 5^x [#permalink] New post 20 Feb 2014, 14:02
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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

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