Algebra or Plug Ins which approach is better

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by Uva@90 » Fri Oct 18, 2013 9:01 pm
[email protected] wrote:If k is a positive integer, is k a prime number?

(1) No integers between 2 and square root of k, inclusive divides k evenly.
(2) No integers between 2 and k/2 divides k evenly, and k is greater than 5.

Ans-D
Hi Shibzriz,

Before jumping into the question, here is the property of the Prime Number,
K is a prime number if no integer between 2 ans square root of k (Inclusive) is a factor of k.
Which is worth remembering it, which is often tested in GMAT(I have seen many times in same forum).

Coming to the question,

Statement 1:
It says the above property so Statement 1 is sufficient to say it is Prime Number.

Statement 2:
No integer between 2 and k/2 divides evenly,

So, no integer divides between 2 and square root of K (Same as statement 1)

hence it is Sufficient.

OA is D

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Uva.
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by [email protected] » Sat Oct 19, 2013 11:49 am
Hi shibsriz,

I'm a big fan of Number Properties and patterns (since both appear on the GMAT repeatedly). However, you DO NOT have time to stare at the screen, trying to figure out IF there's a pattern to be found. As such, sometimes you have to "jump in" and TEST VALUES.

This DS prompt gives us a lot of info to work with. We're told that K is a POSITIVE INTEGER. We're asked if it's prime. This is a Yes/No question.

Fact 1: K is NOT divisible by anything from 2 to (rootK), inclusive.

Since K is NOT divisible by 2, that means that K CANNOT be EVEN.
We CANNOT use K = 1, since root1 divides into 1.
If K = 3, nothing from 2 to root3 divides evenly into 3, so 3 IS allowed. The answer to the question is YES
If K = 5, nothing from 2 to root5 divides evenly into 5, so 5 IS allowed. The answer to the question is YES
We CANNOT use K = 9, since root9 = 3 and 3 divides into 9.
We CANNOT use K = 15, since 3 divides into 15 (and 3 is between 2 and root15).

It appears that the only numbers that "fit" Fact 1 are PRIMES. This is consistent.
Fact 1 is SUFFICIENT

Fact 2: K is NOT divisible by anything from 2 to K/2, inclusive AND K > 5

Since K is NOT divisible by 2, that means that K CANNOT be EVEN
If K = 7, nothing from 2 to 7/2 divides into 7, so 7 IS allowed. The answer to the question is YES
We CANNOT use K = 9, since 3 divides into 9 (and 3 is between 2 and 9/2).
If K = 11, nothing from 2 to 11/2 divides into 11, so 11 IS allowed. The answer to the question is YES
We CANNOT use K = 15, since 3 divides into 15 (and 3 is between 2 and 15/2).

It appears that we have a similar pattern here; the only numbers that "fit" Fact 2 are PRIMES. This is consistent.
Fact 2 is SUFFICIENT.

Final Answer: D

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