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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]

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New post 01 Sep 2019, 15:55
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


I don't know if it's a very "GMAT-like" question ... maybe at 800 level? It's not immediately obvious how best to manipulate and brute force is too time consuming as well, I started by the same reasoning method EMPOWERgmatRichC showed (you can quickly plug in x=1,y=1 to see that x>y) but abandoned it because it's way too much to keep straight and honestly I doubt many people besides Q51 geniuses could reason through that in any sensible amount of time under test conditions... If I saw a question like this on the test I would just guess and move on.

Anyways, a bit of a different solution to those above:

\((5^x)-(5^y)=(2^{y-1})*(5^{x-1})\)

\((5^x)-(2^{y-1})*(5^{x-1})=(5^y)\)

\((5^{x-1})[(5^1)-(2^{y-1})]=(5^y)\)

Since we have powers of 5 on both sides the the only way for LHS to be a power of 5 is if the expression in brackets is = 1
Therefore, \([(5)-(2^{y-1})] = [5-2^{3-1}] = [5-4] = 1\)

We can see that y=3, and now we have \((5^{x-1})[1]=(5^3)\) so \(x-1 = 3\)

Now we can see that x = 4 ... 4*3 = 12, E)
GMAT Club Bot
If x and y are positive integers and 5^x   [#permalink] 01 Sep 2019, 15:55

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