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Re: If x and y are positive integers and [#permalink]
Expert Reply
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 [#permalink]
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Another approach

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 [#permalink]
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I hope now is clear why we cannot have a value >5

regards
Prep Club for GRE Bot
Re: If x and y are positive integers and [#permalink]
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