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If x and y are positive integers and 5^x [#permalink]
06 Apr 2012, 13:01
10
This post received KUDOS
Expert's post
7
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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 \(x\):
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
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This post received KUDOS
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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. _________________
Re: If x and y are positive integers and 5^x - 5^y = 2^(y-1)*5^x [#permalink]
19 Feb 2014, 21:50
5
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.)
If x and y are positive integers and 5^x [#permalink]
19 Aug 2015, 06:30
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
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