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If k is a positive integer, is k a prime number? [#permalink]

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19 Jun 2010, 11:37

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If k is a positive integer, is k a prime number?

(1) No integers between 2 and \(\sqrt{k}\), inclusive divides k evenly. (2) No integers between 2 and k/2 inclusive divides k evenly, and k is greater than 5.

If k is a positive integer, is k a prime number? [#permalink]

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19 Jun 2010, 12:35

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If k is a positive integer, is k a prime number?

Given: \(k=integer>0\). Question: \(k=prime\)?

A prime number is a positive integer with exactly two distinct divisors: 1 and itself.

(1) No integer between 2 and \(\sqrt{k}\) inclusive divides k evenly --> let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\). As given that \(k\) has no factor between 2 and \(\sqrt{k}\) inclusive, then both factors \(a\) and \(b\) must be more than \(\sqrt{k}\). But it's not possible, as the product of two positive integers more than \(\sqrt{k}\) will yield an integer more than \(k\) (\(ab>k\)). Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient.

(2) No integers between 2 and \(\frac{k}{2}\) inclusive divides k evenly, and k is greater than 5 --> the same here : let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\) --> \(k=ab\geq{\frac{k^2}{4}}\) (as both \(a\) and \(b\) are more than or equal to \(\frac{k}{2}\), then their product \(ab\), which is \(k\), must be more than or equal to \(\frac{k}{2}*\frac{k}{2}\)) --> \(4k\geq{k^2}\) --> \(k(4-k)\geq{0}\). But this inequality cannot be true as \(4-k\) will be negative (as given \(k>5\)) and \(k\) is positive so \(k(4-k)\) must be negative not positive or zero. Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient.

Answer: D.

P.S. The first statement is basically the way of checking whether some # is a prime:

Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers (primes) smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq \sqrt{n}\). Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.

x divides k evenly, just means that x is a factor of k, divides k without leaving a remainder. For example, 5 divides 15 evenly --> 15/5=3. _________________

i. k is odd (coz 2 does not divide k). Second test: Let number be 81 and 83. Try few more numbers if you need... only in case of prime numbers condition (i) is true. Hence, SUFFICIENT.

ii. As done in (i), similarly for (ii), k has to be odd (since it does not divide by 2). Second test: Try 9, 11. The condition (ii) is true only in case of prime numbers! Hence, SUFFICIENT.

Re: k is a positive number. Is K a prime number? 1) No integers [#permalink]

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02 Dec 2011, 22:17

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zura wrote:

don't we need word "inclusive" in second statement? that's consider k=6 - no integer between 2 and 3 divides 6 evenly..

Yes, I would like the word 'inclusive' if that is what they mean, especially since they have specifically mentioned 'inclusive' in the first statement. It makes me wonder if they are trying to trick me with a 'ha ha, there was no inclusive in the second statement.'

As for the actual question, it is very straight forward. The following link discusses the theory of factors which includes that a prime number does not have factors in the range 2 to \(\sqrt{PrimeNumber}\) http://www.veritasprep.com/blog/2010/12 ... t-squares/

In case of statement 2, we know that for all n > 5, \(\sqrt{n}\) < n/2 so it basically breaks down and becomes statement 1. _________________

Re: k is a positive number. Is K a prime number? 1) No integers [#permalink]

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06 Jan 2012, 19:32

Good one, got that one correct!

1. The only way integers between 2 and sqrt(k) will be divisible by k is if K is not prime. Try 2, sqrt(9). Now try 2, sqrt(5) -> 5 is not divisible by any integers between 2 and 2.25 (really there is only one integer = 2). Suff. 2. The only way integers between 2 and k/2 will be divisible by k is if K is not prime. try 2 and 6/2. Now try 2, 5/2 and 2, 7/2. Remember integers only. Suff.

D. _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

k is a positive integer. Is k a prime number? [#permalink]

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02 Jun 2013, 14:28

From Barron's passkey to the GMAT Third Edition: Page 320 question #5 k is a positive integer. Is k a prime number?

1) No integer between 2 and \(\sqrt{k}\) inclusive divides k evenly. 2) No integer between 2 and \(\frac{k}{2}\) inclusive divides k evenly, and k is greater than 5.

Statement 1 alone is sufficient since if k is not a prime then k=(m)(n) where m and n must be integers less than k. But this means either m or n must be less than or equal to \(\sqrt{k}\) since if m and n are both larger than \(\sqrt{k}\), (m)(n) is larger than (\(\sqrt{k}\))(\(\sqrt{k}\)) or k. So statement 1 implies k is a prime.

Statement 2 alone is also sufficient, since if k=(m)(n) and m and n are both larger than \(\frac{k}{2}\), then (m)(n) is greater than \(\frac{k^2}{2}\); but \(\frac{k^2}{2}\) is greater than k when k is larger than 5. Therefore, if no integer between 2 and \(\frac{k}{2}\) inclusive divides k evenly, then k is a prime.

Re: If k is a positive integer, is k a prime number? [#permalink]

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03 Sep 2013, 03:22

For sure statment2 requires "Inclusive". E.g) K=8, then between 2 & K/2. There is no interger between 2 & 4 that divides 8.Similary,K=7,then between 2 & K/2 there is no integer that divides K. Hence we require inclusive

Re: If k is a positive integer, is k a prime number? [#permalink]

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13 Nov 2013, 12:54

Hi Bunuel

Fir this question if we consider K to be 7, then K/2 is 3.5. The integers between 2 and 3.5 is only 3 and 3 does not divide 3.5 evenly. Secondly if we take K to be 8, K/2 is 4. Integers between 2 and 4 is 3 only, again 4 is divisible by 3 evenly. Hence, second statement does not show anything. I marked A. Please help!!

Re: If k is a positive integer, is k a prime number? [#permalink]

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13 Nov 2013, 12:58

Expert's post

noorshergill wrote:

Hi Bunuel

Fir this question if we consider K to be 7, then K/2 is 3.5. The integers between 2 and 3.5 is only 3 and 3 does not divide 3.5 evenly. Secondly if we take K to be 8, K/2 is 4. Integers between 2 and 4 is 3 only, again 4 is divisible by 3 evenly. Hence, second statement does not show anything. I marked A. Please help!!

Re: If k is a positive integer, is k a prime number? [#permalink]

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15 Nov 2013, 06:08

Hi Bunuel,

I have read your explanation regarding statement 2, and then re-read it again and again and again.. but I still don't see the answer..

k=ab is pretty clear, but what confuses me is that k=ab>=(k^2)/4.

For example, let's look at a number 100. Then K/2 in this case is 50. If we take two integers a and b which multiply out to give a 100 then a and b, individually, cannot be greater than k/2 or 50 but less...

Re: If k is a positive integer, is k a prime number? [#permalink]

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15 Nov 2013, 08:01

Expert's post

bluecatie1 wrote:

Hi Bunuel,

I have read your explanation regarding statement 2, and then re-read it again and again and again.. but I still don't see the answer..

k=ab is pretty clear, but what confuses me is that k=ab>=(k^2)/4.

For example, let's look at a number 100. Then K/2 in this case is 50. If we take two integers a and b which multiply out to give a 100 then a and b, individually, cannot be greater than k/2 or 50 but less...

Read the highlighted part: (2) No integers between 2 and \(\frac{k}{2}\) inclusive divides k evenly, and k is greater than 5 --> the same here : let's assume \(k\) is not a prime, then there must be some integers \(a\) and \(b\) (\(1<a<k\) and \(1<a<k\)), a factors of \(k\), for which \(ab=k\) --> \(k=ab\geq{\frac{k^2}{4}}\) (as both \(a\) and \(b\) are more than or equal to \(\frac{k}{2}\), then their product \(ab\), which is \(k\), must be more than or equal to \(\frac{k}{2}*\frac{k}{2}=\frac{k^2}{4}\)) --> \(4k\geq{k^2}\) --> \(k(4-k)\geq{0}\). But this inequality cannot be true as \(4-k\) will be negative (as given \(k>5\)) and \(k\) is positive so \(k(4-k)\) must be negative not positive or zero. Hence our assumption that \(k\) is not a prime is not true --> \(k\) is a prime. Sufficient. _________________

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