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DS - Is n be divisible by 4

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DS - Is n be divisible by 4

by zoe » Tue Apr 12, 2016 12:08 am
Dear All

Here is a DS - Is n be divisible by 4
As the attached picture.

My answer is B, I disagree with statement #1 is sufficient,
cuz I have an example:
if N square = 64, obviously is divisible by 8, than n square root = 4√2￣, obviously, it is not divisible by 4, so my answer is insufficient.

any wrong here?
please help
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by [email protected] » Tue Apr 12, 2016 10:40 am
Hi zoe,

Can you post the full question (or type it out)? The image file that you attached doesn't link to an image.

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by zoe » Tue Apr 12, 2016 6:23 pm
[email protected] wrote:Hi zoe,

Can you post the full question (or type it out)? The image file that you attached doesn't link to an image.

GMAT assassins aren't born, they're made,
Rich
Ok,
I didn't input the whole question yesterday because I did not know how to input some math marks.
Maybe input the whole question and add a comment will be better,

A comment:
n2 appearing in the following question means n square, n1/2 appearing in the following question means n square root

please read the following question:

Is positive integer n divisible by 4?
(1) n2 is divisible by 8.
(2) n1/2 is a even integer.

the answer is that each statement alone is sufficient,
I have different idea, I think statement #1 is insufficient, because:
if given n2 = 64, 64 is obviously divisible by 8, and n1/2 = 4√2￣, it is not divisible by 4,
if given n2 = 16, 16 can be obviously by 4,
based on these two given examples above, whether statement #1 is sufficient is depending on the value of n2,
thus, I think statement #1 is insufficient.

please help if I made a mistake.

thanks a lot.
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by Matt@VeritasPrep » Tue Apr 12, 2016 6:37 pm
Zoe, when considering S1 on its own, you don't need to consider S2.

That said, if you know that n² = 8 * something, you have

n * n = 2 * 2 * 2 * something

Since each n is the same, each n must have the same number of 2s in it. We can't have one n = 2 and the other n = 2 * 2; then we'd have different values for the same letter!

So each n must = 2 * 2 * (whatever else), and we know n = 4 * something. That might make √n an even integer and it might not, but we don't really care: we've answered what we were asked.("Yes, n IS divisible by 4.")
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by zoe » Tue Apr 12, 2016 8:30 pm
Matt@VeritasPrep wrote:Zoe, when considering S1 on its own, you don't need to consider S2.
Zoe wrote:if given n2 = 64, 64 is obviously divisible by 8, and n = 4√2￣ (here should be n, and I input n1/2 mistakenly ), it is not divisible by 4,
Dear Matt,
Thanks for your solution.

first, I input n1/2 mistakenly, it should be n = 4√2￣
I think I still need your explanation for my given value , n2= 64, n = 4√2￣, cannot be divisible by 4,
obviously, here is a value that shows the statement #1 is insufficient.

Please help.
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by 800_or_bust » Wed Apr 13, 2016 12:27 pm
Matt@VeritasPrep wrote:Zoe, when considering S1 on its own, you don't need to consider S2.

That said, if you know that n² = 8 * something, you have

n * n = 2 * 2 * 2 * something

Since each n is the same, each n must have the same number of 2s in it. We can't have one n = 2 and the other n = 2 * 2; then we'd have different values for the same letter!

So each n must = 2 * 2 * (whatever else), and we know n = 4 * something. That might make √n an even integer and it might not, but we don't really care: we've answered what we were asked.("Yes, n IS divisible by 4.")
Yes, and we already know from the original prompt that n must be a positive integer. That removes the possibility where n^2=8, which would satisfy (1) but in that case n would equal 2*(2)^(0.5), which conflicts with the requirement that n be a positive integer.

So I think it's important to note the proof you gave works because n must be an integer. Therefore, n must contain at least two factors of two. If we weren't constrained by that, (1) alone wouldn't be sufficient. Am I correct in that assessment?
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by zoe » Wed Apr 13, 2016 5:36 pm
800_or_bust wrote: Yes, and we already know from the original prompt that n must be a positive integer. That removes the possibility where n^2=8, which would satisfy (1) but in that case n would equal 2*(2)^(0.5), which conflicts with the requirement that n be a positive integer.

So I think it's important to note the proof you gave works because n must be an integer. Therefore, n must contain at least two factors of two. If we weren't constrained by that, (1) alone wouldn't be sufficient. Am I correct in that assessment?

Thanks so much.
I got it.

Have a nice day.
>_~
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by Matt@VeritasPrep » Fri Apr 15, 2016 1:02 pm
800_or_bust wrote:So I think it's important to note the proof you gave works because n must be an integer. Therefore, n must contain at least two factors of two. If we weren't constrained by that, (1) alone wouldn't be sufficient. Am I correct in that assessment?
Yup, this is correct. Since we're told in the prompt that n is an integer, we only need to consider integer solutions, but if we were considering all numbers, we'd have

n² = 8k, where k is some integer we don't care about

n = 2√(2k)

So if n is an integer, then it's an even integer, but whether it's an integer at all depends on the value of k.
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by Matt@VeritasPrep » Fri Apr 15, 2016 1:04 pm
zoe wrote:
Matt@VeritasPrep wrote:Zoe, when considering S1 on its own, you don't need to consider S2.
Zoe wrote:if given n2 = 64, 64 is obviously divisible by 8, and n = 4√2￣ (here should be n, and I input n1/2 mistakenly ), it is not divisible by 4,
Dear Matt,
Thanks for your solution.

first, I input n1/2 mistakenly, it should be n = 4√2￣
I think I still need your explanation for my given value , n2= 64, n = 4√2￣, cannot be divisible by 4,
obviously, here is a value that shows the statement #1 is insufficient.

Please help.
If n² = 64, then n = 8, so n would be divisible by 4.
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