D S For Number Properties.

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D S For Number Properties.

by gaurav7infy » Thu Feb 07, 2013 6:14 am
x,y, and z are integers, is x even?
(1) 1O^x = (4^x)(5^z)
(2) 3^(x+5) = 27^(y+1)

Please Help.

Thanks.
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by Ian Stewart » Thu Feb 07, 2013 6:58 am
There's something wrong with the question - where is it from? Either it's badly designed, or Statement 1 is meant to read something more like: 10^x = (4^z)(5^x).

As written, it's impossible for both statements to be true. From Statement 1 we have:

2^x * 5^x = 2^(2x) * 5^z

Our exponents are integers, so they need to be the same on both sides of this equation. So, looking at the powers on the 2s on the left and right sides, x = 2x must be true, and the only possibility is that x=0. So Statement 1 is sufficient.

But if you look at Statement 2, x cannot be 0, so that makes the two statements inconsistent, which is why the question is either flawed or incorrectly transcribed. In any case, Statement 2 is not sufficient; we find that

3^(x+5) = 3^(3y + 3)

so x+5 = 3y+3, and x = 3y -2. Since 3y-2 might be even or might be odd, we don't know if x is even.
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by gaurav7infy » Thu Feb 07, 2013 4:53 pm
Thanks Ian for reply.
very Sorry :( my typo mistake.
correct que is as below.

x, y, and z are integers, is x even?
(1) 1O^x = (4^y)(5^z)
(2) 3^(x+5) = 27^(y+1)

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by Ian Stewart » Fri Feb 08, 2013 12:16 am
gaurav7infy wrote:Thanks Ian for reply.
very Sorry :( my typo mistake.
correct que is as below.

x, y, and z are integers, is x even?
(1) 1O^x = (4^y)(5^z)
(2) 3^(x+5) = 27^(y+1)
That question makes more sense. :)

From Statement 1 we know:

10^x = (4^y)(5^z)
(2*5)^x = (2^2)^y (5^z)
2^x * 5^x = 2^(2y) * 5^z

and because we have prime bases and the exponents must be integers, the powers on the 2 on either side of the equation must be equal, and the powers on the 5 on either side must be equal. So looking at the powers on the 2, we see that x = 2y, and x is therefore even. So Statement 1 is sufficient. Statement 2 is not sufficient, as discussed above, so the answer is A.
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