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Kaplan DS 43- Prime number

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by rishimaharaj » Tue Aug 02, 2011 6:04 pm
Hello inertia2010,

Here is my quick attempt...
Question Stem Rephrased:
Is z prime?
z != prime if it is a multiple of 2. (Edit: '!=' is used in programming languages to mean NOT EQUAL TO)
z != prime it is divisible by any other factors.
Does z have any factors other than itself and 1?

1. 15! < z
Rephrased this says
-Big EVEN # < z.
-z can be either even or odd and goes all the way up to infinity.
-therefore, z can either be a prime, or a normal #.
-INSUFFICIENT.

2. 17!+2 <= z <= 17!+17 (Edit: '<=' is used in programming languages to mean LESS THAN OR EQUAL TO)
Rephrased this says
-Big EVEN # <= z <= Big Multiple of 17.
-so z can be 17!+2 (EVEN, not prime), 17!+17 (Mult of 17, not prime), and all the numbers in between:
17!+3,4,5,6,7,8,9,10,11,12,13,14,15,and 16...
Since 17!+2 is even, 17!+4,6,8,10,12,14,and 16 are even, so they are all not prime.

This can be sufficient if we can prove that 17!+3,5,7,9,11,13,and 15 are non prime. If any of them ARE prime, then this statement will be insufficient.

Note: if this is INSUFFICIENT, then the answer will be E, because combining both statements will just result in whatever the answer is for statement two, since both statements are inequalities and statement two has the tightest restrictions (smallest range of values).

At this point we're a little stuck, because, hell, 17! is a huge number. So what to do?

We can notice that 17! contains each factor from 2 through 17, so this means that each number in 17! + that number will be a multiple of that number.
17! + 2 will be a multiple of 2.
17! + 3 will be a multiple of 3.
...
17! + 15 will be a multiple of 15.
17! + 16 will be a multiple of 16.

Hence, since all of those values are non primes, Statement 2 lets us definitively answer the question: is z prime? NO, it is not. Statement 2 is SUFFICIENT.

The answer is B.

I'm not an expert, and there might be a quicker way to solve this, or realize the "trick." I didn't figure it out until I got stuck, but then, it just hit me. :-)

Hope this helps. If you still have any doubts, please let me know.


Thanks,
--Rishi
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by gmatboost » Tue Aug 02, 2011 8:42 pm
You're right. The key part of your answer is:
rishimaharaj wrote: We can notice that 17! contains each factor from 2 through 17, so this means that each number in 17! + that number will be a multiple of that number.
17! + 2 will be a multiple of 2.
17! + 3 will be a multiple of 3.
...
17! + 15 will be a multiple of 15.
17! + 16 will be a multiple of 16.
Beyond that, there aren't really any shortcuts. There is an almost identical question in the OG that uses 13! instead of 17!
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