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Integer Properties.

This topic has 13 expert replies and 1 member reply
kamalakarthi Senior | Next Rank: 100 Posts Default Avatar
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Integer Properties.

Post Wed Aug 02, 2017 3:16 am
GMAT Gurus,

In this DS question, I came to conclusion that Statement 1 is not sufficient. But, it turns out to be wrong. Can you please help on what am I missing in the big picture.

My work: Since ab = 2, either both a and b are + ve or -ve

in case of +ve, It could be 2 and 1 or 4 and 1/2 if it is 2 and 1 then [a] + [b] = 1 as a would be 1 and b would be 0 but this is not true if a was 4 and b was 1/2. Same for -ve as well. So, I thought statement A is not sufficient.

Statement 2, I was able to conclude 2 is not sufficient.
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Post Sun Aug 06, 2017 10:04 pm
Another way we could get there:

ab = 2

a = 2/b

With that, we have [a] + [b] => [2/b] + [b].

We also know that a and b must be nonnegative integers. (If one of them is negative, then the other must be negative, since a = 2/b, and if they're both negative, then [a] and [b] are both < 0.)

Now let's suppose that [2/b] + [b] = 1. Since both [2/b] and [b] are nonnegative integers, the only way we can get a sum of 1 is if one of these = 1 and the other = 0.

Let's take the first case, [b] = 1 and [2/b] = 0. If that's the case, then 1 ≤ b < 2 and 0 < 2/b < 1. That second inequality simplifies as 2 < b. With that in mind, we must have b > 2 ... but if b > 2, then [b] ≥ 2, and we saw above that [b] = 1. That's a contradiction, so [b] = 1 and [2/b] = 0 is impossible.

The second case ([b] = 0 and [2/b] = 1) works much the same way.

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Post Sun Aug 06, 2017 9:57 pm
kamalakarthi wrote:
in case of +ve, It could be 2 and 1 or 4 and 1/2 if it is 2 and 1 then [a] + [b] = 1 as a would be 1 and b would be 0 but this is not true if a was 4 and b was 1/2. Same for -ve as well. So, I thought statement A is not sufficient.
Let's assume that both a and b are positive. (If they're both negative, then obviously their floors can't sum to 1.)

If a > 0, b > 0, and ab = 2, then AT LEAST ONE of a and b must be ≥ √2. (If they're both < √2, then their product is less than √2 * √2.) If one of them is > √2, then the other must be < √2, or else their product will be > 2.

But whichever one is ≥ √2 must be at least [√2], or at least 1! Let's say that one is a. Then we've got 1 + [b], where b = 2/a. The only way that [b] = 0 is if [2/a] = 0, or if 0 < 2/a < 1. But multiplying that inequality by a, we get 0 < 2 < a, or 2 < a, which can't be, since we can't have both terms be greater than √2.

So we can't get a sum of 1, no matter what.

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Post Sat Aug 19, 2017 5:18 am
Matt@VeritasPrep wrote:
Brent@GMATPrepNow wrote:
Matt@VeritasPrep wrote:
Brent@GMATPrepNow wrote:
kamalakarthi wrote:
The source of this question is from gmatclub.com quant tests.
Too late. The police are already on their way Smile
But they didn't read him his copyrights, so he's off the hook Very Happy
Perhaps, but you're not off the hook for that pun!!!
Brent, you are no pun at all :p
I happun to believe I'm very pun Smile

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Post Fri Aug 18, 2017 1:20 pm
Brent@GMATPrepNow wrote:
Matt@VeritasPrep wrote:
Brent@GMATPrepNow wrote:
kamalakarthi wrote:
The source of this question is from gmatclub.com quant tests.
Too late. The police are already on their way Smile
But they didn't read him his copyrights, so he's off the hook Very Happy
Perhaps, but you're not off the hook for that pun!!!
Brent, you are no pun at all :p

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Post Mon Aug 07, 2017 6:19 am
Matt@VeritasPrep wrote:
Brent@GMATPrepNow wrote:
kamalakarthi wrote:
The source of this question is from gmatclub.com quant tests.
Too late. The police are already on their way Smile
But they didn't read him his copyrights, so he's off the hook Very Happy
Perhaps, but you're not off the hook for that pun!!!

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Post Sun Aug 06, 2017 10:09 pm
One last way to get there (and to get a feel for what's happening) is trying numbers.

A simple case if a = √2 and b = √2. That gives us [a] = 1 and [b] = 1 and [a] + [b] = 2. OK, too big, but let's try monkeying around with the numbers. For simplicity's sake, we'll make a > b.

If a ≥ 1.5, then [a] = 2, and [a] + [b] > 1, so we can't have that either. With that in mind, we MUST have 1.5 > a ≥ √2, so a is in a pretty tight band between about 1.415 and 1.5. But if a's in that range, then b is somewhere between √2 and 2/1.5, or somewhere between 1.333... and about 1.415. With that in mind, our ranges are

1.5 > a ≥ √2, or [a] = 1

√2 ≥ b > 1.333..., or [b] = 1

So by trying to keep a in the right range, we can't get any smaller than [a] + [b] = 1 + 1.

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Post Sun Aug 06, 2017 9:52 pm
Brent@GMATPrepNow wrote:
kamalakarthi wrote:
The source of this question is from gmatclub.com quant tests.
Too late. The police are already on their way Smile
But they didn't read him his copyrights, so he's off the hook Very Happy

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Post Wed Aug 02, 2017 11:34 am
Brent@GMATPrepNow wrote:
kamalakarthi wrote:
The source of this question is from gmatclub.com quant tests.
Too late. The police are already on their way Smile
Very Happy

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Post Wed Aug 02, 2017 11:19 am
kamalakarthi wrote:
The source of this question is from gmatclub.com quant tests.
Too late. The police are already on their way Smile

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kamalakarthi Senior | Next Rank: 100 Posts Default Avatar
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Post Wed Aug 02, 2017 10:43 am
Experts, Thanks for the reply. The source of this question is from gmatclub.com quant tests.

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Post Wed Aug 02, 2017 10:26 am

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Post Wed Aug 02, 2017 10:14 am
This formula structure of "greatest integer less than or equal to" or "least integer greater than or equal to" is very common on the GMAT. Here are some other examples to practice with:
http://www.beatthegmat.com/ds-t276593.html#719492
http://www.beatthegmat.com/ds-t281646.html#738122
http://www.beatthegmat.com/if-denotes-the-t293923.html#787526

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Post Wed Aug 02, 2017 10:12 am
Please also post the source of this question. It's a copyright violation not to do so!

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Post Wed Aug 02, 2017 10:11 am
Be careful - you are slightly misinterpreting the prompt: [x] denotes the greatest integer less than or equal to x. For any non-integer, we round down to the nearest integer. But whenever x is an integer, then [x] = x.

Use the same testing cases approach you were using before:

(1) ab = 2

Case 1:
a = 2 --> [a] = 2
b = 1 --> [b] = 1
[a] + [b] = 2 + 1 = 3

We get a "no" answer to the question. So can we get a "yes" answer to the question?

Your initial analysis was sound: a product of 2 tells us that we have (positive*positive) or (negative*negative). If both a and b are negative, then [a] and [b] must also be negative, so their sum can never equal 1.

If both positive, then either both a and b equal the square root of 2, or one is greater and the other is smaller than the square root of 2.

Case 2:
a = sqrt(2) = approx. 1.4 --> [a] = 1
b = sqrt(2) --> [b] = 1
[a] + [b] = 2

Again, a "no" answer to the question. Since this is the closest together that we can get the two values, it is the minimum possible value of [a] + [b]. Therefore, there is no possible way that [a] + [b] = 1. We have sufficient information to know that the answer is definitively "no."

Does that help?

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