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by tlt2372 » Wed Oct 27, 2010 5:01 pm
If (2^x)(3^y) = 288, where x and y are positive integers, then (2^(x-1))(3^(y-2))=?

A)16
B)24
C)48
D)96
E)144

I know you can just pick numbers and get lucky, but is there a shortcut?

Thanks

OA: A
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by Taniuca » Wed Oct 27, 2010 5:26 pm
PROCESS

2^x. 3^y = 288 find all the factors for 288= 2*2*2*2*2*3*3 that is the same as saying 2^5*3^2

2^x * 3^y = 2^5*3^2
so, x=5 and y=2

Plug the numbers in your second equation

2^ (x-1) * 2 (y-2) =
2^ (5-1) * 2 (2-2) =
2^4 * 2^0=
16 *1= 16
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by tlt2372 » Wed Oct 27, 2010 5:38 pm
Thanks!
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by limestone » Wed Oct 27, 2010 8:50 pm
Hi,

I have the general rule for this.

If you have :

a^x * b^y = k, and you want to find a^(x-n) * b ^ (y-m), you can divide k to (a^n * b^m)

In this case : 2^x * 3^y = 288, then you wanna find 2^(x-1)* 3^(y-2):

Then 2^(x-1)* 3^(y-2) = 288/ (2^1 * 3^2) = 288/18 = 16

Hope this helps.
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by Rahul@gurome » Wed Oct 27, 2010 9:08 pm
tlt2372 wrote:If (2^x)(3^y) = 288, where x and y are positive integers, then (2^(x-1))(3^(y-2))=?

A)16
B)24
C)48
D)96
E)144

I know you can just pick numbers and get lucky, but is there a shortcut?

Thanks

OA: A
Yes, there exists a method to solve this kind of problems.
For variable x and constant n,
(1) a^(x + n) = (a^x)*(a^n)
(2) a^(x - n) = (a^x)/(a^n)

Thus,
[2^(x - 1)][3^(y - 2)] = [(2^x)/(2)][(3^y)/(3^2)] = (2^x)(3^y )/(2*9) = 288/18 =16

The correct answer is A.

Note: Plugging the options work for this particular problem. But if the numbers were large or the question made more complicated, plugging options will become cumbersome and too some extent impossible.
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