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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

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.
_________________

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]

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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.)

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, 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
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Any and all kudos are greatly appreciated. Thank you.

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

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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.
_________________

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

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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]

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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]

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26 Aug 2016, 18:47

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

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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.

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

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