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)