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Data Sufficiency

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by parallel_chase » Mon Aug 25, 2008 5:50 am
x odd number
Statement I

x^x=y^2

x=1, y=1

Sufficient.

Statement II

y is an integer. Insufficient.

Hence A.
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by 4meonly » Mon Aug 25, 2008 7:11 am
parallel_chase wrote:x odd number
Statement I

x^x=y^2

x=1, y=1

Sufficient.

Statement II

y is an integer. Insufficient.

Hence A.
I think E

(1)
x^x also may be 9^9
9^9=387420489
sqrt (9^9)=sqrt 387420489=19683
it can be expressed
3*9^4*3*9^4
so x=1 is not one answer
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by jeffxujian » Tue Aug 26, 2008 12:03 am
I think C

but I was definitely inspired by earlier two answers.

1. X can be anything, because Y is not specified as if it is a interger or not, for example:

3^3 = <3sqrt3>^2, so 3 is another valid answer besides 1

if we combine 1 and 2, we will come up with X=1 and it is sufficient to answer to question. Please correct if i am wrong. Thanks!
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by pepeprepa » Tue Aug 26, 2008 2:17 am
I agree with C

1)
1^1=y^2
3^3=y^2 is possible if y=sqrt(27) or y=-sqrt(27)
x has different possibilities

2)
Useless

1)and2)
We only have 1
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by 4meonly » Tue Aug 26, 2008 7:06 am
4meonly wrote:
parallel_chase wrote:x odd number
Statement I

x^x=y^2

x=1, y=1

Sufficient.

Statement II

y is an integer. Insufficient.

Hence A.
I think E

(1)
x^x also may be 9^9
9^9=387420489
sqrt (9^9)=sqrt 387420489=19683
it can be expressed
3*9^4*3*9^4
so x=1 is not one answer
Guys, where is my mistake?

&#1093; can be as 1 as 9, and they are both odd!

Answer E.

Where is my mistake?
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by rishi235 » Tue Aug 26, 2008 8:40 am
@ 4meonly...

I guess ur assumption x=3*9^4 does not satisfy the 2nd condition

Now x^x = y^2
=> y = x^(x/2)
=> (3*9^4) ^ [(3*9^4)/2]
=> Now 3*9^4 / 2 is not an integer
Hence Integer ^ Non-integer does not = Integer...so
(3*9^4) ^ [(3*9^4)/2] is not an integer...
Hence this does not satisfy the 2nd condition...
Only x=1 & y=1 will satisfy both the condition...
Hence Ans is C...
Correct me if m wrong....
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by Ian Stewart » Tue Aug 26, 2008 10:15 am
4meonly wrote: Guys, where is my mistake?

&#1093; can be as 1 as 9, and they are both odd!

Answer E.

Where is my mistake?
You didn't make a mistake! Your solution is perfect, and the answer should be E. As 4meonly pointed out, x can be 9:

x^x = 9^9 = (3^2)^9 = (3^9)^2

So if x = 9, y would then be 3^9, which is certainly an integer. Indeed, there are many other possible values of x- x could be any square of an odd integer.
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by dalwow » Thu Aug 28, 2008 9:53 am
Ian-

Is there a faster way to approach this problem? It doesn't seem like we would be expected to work it out for all the digits. Additionally, just multiplying it out for 9^9 and finding the square root of that would take considerable time. What are the <2 minute approaches?
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by Ian Stewart » Thu Aug 28, 2008 2:49 pm
dalwow wrote:Ian-

Is there a faster way to approach this problem? It doesn't seem like we would be expected to work it out for all the digits. Additionally, just multiplying it out for 9^9 and finding the square root of that would take considerable time. What are the <2 minute approaches?
Well, no, there's no need to even choose a number here. The answer is clearly C or E. Using both statements, we only know

x^x = y^2

and that x and y are odd integers. That is, the question is really asking, if x is odd, how can x^x be a perfect square? Is x^x only a perfect square when x = 1? Well, no. If I raise any odd perfect square to any positive integer power x, I'll get another perfect square, so I just need to make x = p^2 for any odd number p. There are thus infinitely many solutions for x.

Algebraically, if x = p^2, then x^x = (p^2)^x = (p^x)^2. So if x = p^2, y will be p^x or -p^x.
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by dalwow » Fri Aug 29, 2008 5:00 am
Thanks. That makes sense. I was just trying to grab the logic behind this question.
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