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DS: Primes #2

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DS: Primes #2

by saintforlife » Sun Sep 02, 2012 6:54 pm
Does there exist an integer d such that x>d>1 and x/d is an integer?
(1) 11!+2≤x≤11!+12
(2) x≥2^5

My answer was A. Is that that correct answer?

My thought process was:

(2) x≥2^5 - Not sufficient => Because there are many numbers greater than 32 that are both primes and composites. Eliminates B and D.
(1) 11!+2≤x≤11!+12 - Sufficient => Because in the first expression 2 can be factored out between 11! and 2 and in the 2nd expression, 6 can be factored out between 11! and 12.

I've seen problems before where the choice is something like 11!+2≤x≤11!+11, where the 2nd term in the 2nd expression is the same as the factorial (11!+11). But in this problem, the 2nd term is 12 (11!+12). Does this change anything? In my opinion it doesn't, because the 2nd expression (11!+12) is still a multiple of 6. Thoughts?

Also I'd like an expert to comment on this:

Somebody told me that if the (1)st answer choice said: 11!+2<x<11!+12 instead of 11!+2≤x≤11!+12, there is no way to solve this problem. I am thinking it is the "=" to sign that is the difference maker. Can somebody explain if I am on the right track and the exact reasoning behind this assertion?
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by pemdas » Sun Sep 02, 2012 9:20 pm
d<x and d<11!+2 anyways
why can't d be 2 or 4 or 10 and x will be divisible by these numbers still

what's OA?
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by Ian Stewart » Sun Sep 02, 2012 10:47 pm
saintforlife wrote:
Somebody told me that if the (1)st answer choice said: 11!+2<x<11!+12 instead of 11!+2≤x≤11!+12, there is no way to solve this problem.
No, that is not right; it really doesn't matter in this question. Either way, you need to consider all of the integers *between* 11! + 2 and 11! + 12. In your solution, you seem only to have considered the smallest and largest numbers in the range, but you certainly also need to consider all of the numbers in the middle. So for Statement 1 to be sufficient, we need to be sure that none of the numbers:

11! + 2, 11! + 3, 11! + 4, ... 11! + 11, 11! + 12

is prime. We can tell quite quickly that 11! + 2 and 11! + 12 are not prime, because they're both even numbers; 11! is even, so 11! + 2 and 11! + 12 are each the sum of two even numbers, and are thus even and not prime. That's why it doesn't really make any difference if the question is written with '<' inequalities or '<' inequalities; the extreme values in the interval are not the hard part of the problem.

This question is testing an elementary, but crucially important, principle of divisibility: if you add or subtract two multiples of some number d, you must always get a multiple of d. So, for example, if you add two multiples of 7, you will always get a multiple of 7. This is true because we could write our two multiples of 7 as '7m' and '7n', and when we add them, we can factor out a seven: 7m + 7n = 7(m + n), which is clearly a multiple of 7.

Now, notice that 11! = 11*10*9*8*7*6*5*4*3*2*1 is a multiple of *every* integer up to 11. So 11! + 7, say, is just the sum of two multiples of 7, and must be a multiple of 7. So it can't be prime. Similarly, 11! + 3 is the sum of two multiples of 3, so it must be a multiple of 3, and is not prime. Since we could use the same logic for any number between 11! + 2 and 11! + 11 inclusive, none of the numbers between 11! + 2 and 11! + 11 inclusive can be prime, and Statement 1 is sufficient. More generally, if n is an integer greater than 1, there are never any primes between n! + 2 and n! + n inclusive.

It's possible to change the question so that it becomes impossible to answer with pen and paper. If a large number has no obvious divisors, it can take an enormous amount of time to work out if that number is prime. So it would be completely impossible to work out in 2 minutes with just pen and paper whether 11! + 1 is prime (using the logic above, it's a sum of two multiples of 1, so it must be a multiple of 1, but that isn't very useful - everything is a multiple of 1). And because it would be impossible to figure out if that number is prime in 2 minutes (or even 2 hours) with just pen and paper, the GMAT simply can't ask you about it. But that's the kind of question your friend might have been talking about when he or she said that there might be no practical way to solve this type of problem if the range in the question were different.
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by mba404 » Sun Sep 02, 2012 11:04 pm
Yes.. Indeed answer is A.

Statement 1 Alone: 11!+2≤x≤11!+12

For all the numbers 11!+n where n≤11 , n divides 11!+n (11! is divisible by n and n is divisible by n so 11!+n is also divisible by n) -> at least one d exists for each of these numbers

For 11!+12 we have number 2,3,4 and 6 that divide both 11! and 12 hence at least one d exists for 11!+12

Statement1 is sufficient

Statement 2 Alone: x≥2^5

if x= 2^5 we can have d=2
if x is any prime number greater then 2^5 we won't have d
Statement2 is Insufficient


IMO A
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by vk_vinayak » Mon Sep 03, 2012 7:18 am
Does there exist an integer d such that x>d>1 and x/d is an integer?
(1) 11!+2≤x≤11!+12
(2) x≥2^5
Why the answer is not D?

The question asks if there exists an integer that satisfies the condition. If we show one such integer, can we not say YES?
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by neelgandham » Mon Sep 03, 2012 7:47 am
vk_vinayak wrote:
Does there exist an integer d such that x>d>1 and x/d is an integer?
(1) 11!+2≤x≤11!+12
(2) x≥2^5
Why the answer is not D?

The question asks if there exists an integer that satisfies the condition. If we show one such integer, can we not say YES?
Case 1: If x = 32, then there exists a d(2,4,8,16)
Case 2: If x = 37, then there is no d(>1) because 37 is a prime number.

Es claro ?
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by vk_vinayak » Mon Sep 03, 2012 7:54 am
neelgandham wrote:
vk_vinayak wrote:
Does there exist an integer d such that x>d>1 and x/d is an integer?
(1) 11!+2≤x≤11!+12
(2) x≥2^5
Why the answer is not D?

The question asks if there exists an integer that satisfies the condition. If we show one such integer, can we not say YES?
Case 1: If x = 32, then there exists a d(2,4,8,16)
Case 2: If x = 37, then there is no d(>1) because 37 is a prime number.

Es claro ?
I was thinking that if for any integer there exists an integer d that satisfies the condition. I get it now. Thanks for the clarification.
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by pemdas » Mon Sep 03, 2012 3:34 pm
neelgandham wrote:
vk_vinayak wrote:
Does there exist an integer d such that x>d>1 and x/d is an integer?
(1) 11!+2≤x≤11!+12
(2) x≥2^5
Why the answer is not D?

The question asks if there exists an integer that satisfies the condition. If we show one such integer, can we not say YES?
Case 1: If x = 32, then there exists a d(2,4,8,16)
Case 2: If x = 37, then there is no d(>1) because 37 is a prime number.

Es claro ?
so the whole question's point is to ask for at least one possibility of d=integer

this is funny problem, I must tell you
not unique value, question asks an integer i.e. one single or we switch on "at least" condition?
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