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Is n an integer? My logic is failing me...

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Is n an integer? My logic is failing me...

by mixpanda » Thu Apr 16, 2009 3:59 pm
From Official Guide 11 practice questions:

Is n an integer?

1.) n squared is an integer
2.) the square root of n is an integer

My thinking:

1.) While considering the listed fact, we have to assume that the fact is true. We must assume that n squared is an integer. The only way that n squared is an integer is if n is also an integer. If n squared is an integer, then n must also be an integer. SUFFICIENT.

2.) in order for the square root of n to be an integer, n must also be an integer. SUFFICIENT.

I would answer D. That isn't the right answer. Explanation?
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Source: — Data Sufficiency |
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by mixpanda » Thu Apr 16, 2009 4:08 pm
Ok, actually...

When I answered the question, I got the correct answer. But a statement in the explanation made me change my mind to D, which is wrong.

I originally thought this: even if n squared is an integer, n could be some impossibly long decimal that could be squared to form an integer.

But the explanation says:

While n squared is an integer, since n squared = n x n, then n squared is an integer if n is an integer; it is unclear whether n is an integer here; NOT sufficient."

But that logic makes no sense! If n squared is an integer as long as n itself is an integer, and we are assuming that n squared IS IN FACT an integer, doesn't that mean n must be an integer? The logic is backwards, but it's still true, right?
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by mixpanda » Thu Apr 16, 2009 4:12 pm
Sorry! Forgot...

Official Answer is B
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by 4seasoncentre » Thu Apr 16, 2009 4:20 pm
For statement (1)

Thinking backwards, let's take the number 4. It is an integer, and it is the square of 2, which is also an integer.
Now lets think of the number 5. It is an integer, and it is the square of sqrt(5). Sqrt(5) is definately not an integer as it is equal to 2.somethingsomething. You don't have to know what it is, just that it has decimals.

This makes statement (1) insufficient, as I have found 2 instances that comply to the statement that yield separate results for the question.

As for statement (2)

No need to think backwards. Take the number 2 square it, you get 4. Take the number 3, square it, you get 9. Take (-1), square it. Take (-3), square it etc etc.

I can't seem to find a value of N where the square root is an integer and N is not an integer. So 2 is sufficient.

I can stop here and answer B.
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by mixpanda » Thu Apr 16, 2009 4:27 pm
So then the explanation given by OG is flawed? Your explanation makes perfect sense, and that's how i originally realized answer B was correct. But when I read that n squared being an integer was dependent on n being an integer, I got confused.

I guess its not true then, correct? n squared can be an integer without n having to be an integer. So the OG's explanation is flawed, but the answer is correct.
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by 4seasoncentre » Thu Apr 16, 2009 4:37 pm
No, I don't see anything inaccurate about the solution, but it sort of answers statement 2, then goes back to statement 1.

Kinda confusing.
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