If x is an integer, does x have a factor n such that 1

Question

If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15

Bunuel · Answer

If x is an integer, does x have a factor n such that 1 < n < x?

(1) x > 3!

(2) 15! + 2 ≤ x ≤ 15! + 15

Show Answer

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VeritasPrepBrian · Answer

I love this question, so I'll chime in even if only to bump it to the top so more people see it!

Hopefully most can see pretty quickly that x > 3! just means x > 6, and that isn't nearly enough to tell us whether it is prime.

Statement 2 is pretty neat, though: 15! + 2 ≤ x ≤ 15! + 15

Think about 15!. 15! is 15*14*13*12*11*10*9*8*7*6*5*4*3*2*1, which means that:

15! is a multiple of 2
15! is a multiple of 3
15! is a multiple of 4
etc....
15! is a multiple of 15

Now think about multiples of 2. Every SECOND number is a multiple of 2: 2, 4, 6, 8, 10. You create multiples of 2 by adding 2 to a previous multiple of 2. If a number is even, adding 2 "keeps it" even.

The same holds for 3. Every THIRD number (3, 6, 9, 12, 15, 18) is a multiple of 3. If you have a multiple of 3 and add 3 to it, it's still a multiple of 3.

This will hold for all of these numbers - because 15! is a multiple of every number between 2 and 15, then adding any number in that range of 2-15 will ensure that the new number remains a multiple of that number, and the new number will not be prime.

Therefore, statement 2 is sufficient, and the correct answer is B.

Now...consider the number 15! + 1. It's too big a number to know offhand whether it's prime, but we do know that it is NOT a multiple of any numbers 2-15. In order to be a multiple of 2, we'd have to add a multiple of 2 to 15!, and 1 breaks us off of that every-second-number cycle. Same for 3 - we'd need to add a multiple of 3 in order to keep 15! on that every-third cycle, so 15! is not a multiple of 3. We can prove that 15! + 1 is not divisible by any numbers between 2 and 15. It's largest prime factor must then be 17 or greater.

Understanding that ideology and being able to determine divisibility of large numbers can be quite helpful on questions that might otherwise seem impossible. Thanks for posting this question!

zisis · Answer

I am able to follow until the point that I have highlighted red.......
tried to work your point my self so decided to find if x for 5!+3<x<5!+8 works with your statement above....

so - 
 5! = 120 
 5!+3 / 3 = 123 / 3 =41 correct
 5!+4 / 4 = 124 / 4 =31  correct
 5!+5 / 5 = 125 / 5 =25  correct
 5!+6 / 6 = 126 / 6 =21 correct
   5!+7 / 7 = 127 / 7 =18.14 INCORRECT !!!!

so...? what am i doing wrong???

Bunuel · Answer

You replaced 15! by 5!. Thus you cannot factor out 7 out of 5!+8 and this is the whole point here.

If you want to check with smaller numbers try  5!+2\leq{x}\leq{5!+5}

zisis · Answer

did it on excel and seems like you are right.....

15! 15!+x (15!+x)/x
1,307,674,368,000.00 1,307,674,368,003.00 435,891,456,001.00
1,307,674,368,000.00 1,307,674,368,004.00 326,918,592,001.00
1,307,674,368,000.00 1,307,674,368,005.00 261,534,873,601.00
1,307,674,368,000.00 1,307,674,368,006.00 217,945,728,001.00
1,307,674,368,000.00 1,307,674,368,007.00 186,810,624,001.00
1,307,674,368,000.00 1,307,674,368,008.00 163,459,296,001.00

now have to go back and understand how it works.....using the example taht I used (which you mentioned I should use smaller numbers), how do you know when x is too big for a and b, when k!+a<x<k!+b
?

Bunuel · Answer

It seems that you don't understand the explanation. No need to check in excel: no integer  x  satisfying  15!+2\leq{x}\leq{15!+15}  will be a prime number.

There are 14 numbers satisfying it:
If  x=15!+2  then we can factor out 2, so  x  would be multiple of 2, thus not a prime;
If  x=15!+3  then we can factor out 3, so  x  would be multiple of 3, thus not a prime;
...

If  x=15!+15  then we can factor out 15, so  x  would be multiple of 15, thus not a prime.

Also  15!+{any multiple of prime less than or equal to 13}  also won't be a prime number as we can factor out this prime (15! has all primes less than or equal to 13).

Now, for  x=15!+1  or  x=15!+17  or  x=15!+19  we can not say for sure whether they are primes or not. In fact they are such a huge numbers that without a computer it's very hard and time consuming to varify their primality.

Hope it's clear.

NGGMAT · Answer

good qs!! and explainaton bunnel

i am stalking you on GMATclub.com

chalmers15 · Answer

Hey Bunuel,

regarding statement 2: Let's say we would add a prime number bigger than 15 to the 15!, e.g. 17. Because it is not possible to factor out 17, the statement is not sufficient, right? Because the result might be prime or not. On the other hand, if we would add 16, we could still factor out 2 and 8 from 15! correct?

Best regards
chalmers15

Bunuel · Answer

You are almost right.

Yes, 15! + 16 is not a prime because we can factor out an integer greater than 1 from 15! + 16.
Yes, 15! + 17 might be a prime as well as not. But 15! + 17 is some specific number, so theoretically we could find out whether it's prime or not. If it were a prime, then  15!+2\leq{x}\leq{15!+17}  would not be sufficient and if it were NOT a prime, then  15!+2\leq{x}\leq{15!+17}  would be sufficient. Turns out that 15! + 17 is NOT a prime number: 15! + 17 = 1307674368017 = 9157×142805981. So, if (2) were  15!+2\leq{x}\leq{15!+17}  it would still be sufficient. But you won't get such number on the GMAT because you'll need a computer to check whether such a large number is a prime (there are some particular values of  x=n!+1  for which we can say whether it's a prime or not with help of Wilson's theorem, but again it's out of the scope of GMAT.).

xiaoxueren · Answer

I have a question, if 15! + 1 is a prime number ?
15! + 1=1(15*14*13*.....*2+1) 
so 15! + 1 is a prime number?

VeritasPrepBrian · Answer

Yeah really good question. There *is* of course an answer to that, but not one you can prove in 2-3 minutes without a calculator so the GMAT couldn't really ask you about that. You *can* prove that it's not divisible by anything between 2 and 15, but you don't have any easy proof for prime numbers greater than 15 (could it be divisible by 17 or 31 or 53? We just don't know).

Consider 5! + 1. 5! is 120 so 5! + 1 = 121. We know that 5! + 1 can't be divisible by 2, 3, 4, or 5, but the factorial setup doesn't give us any insight as to whether it's divisible by 11 (it is).

So with 15! + 1, we can't predict (without a calculator at least) which larger prime factors it might have, so in the world of GMAT we can't prove whether it's prime or not.

sajibmannan · Answer

The question asks if we pick any random integer x, will it have a factor  n  in the format  1<n<x ? Will the integer x have a factor smaller than itself?

Statement 1  says that sometimes it will, and sometimes it won't. This is because sometimes in that range there will be a prime number, and sometimes it won't.

Example:

If x=7, there won't be a factor n in this format  1<n<x  because 7 has only two factors (1, 7).

If x=8, then there WILL be a factor n in this format  1<n<x  because 8 has four factors (1, 2, 4, 8).

Thus, statement 1 is insufficient.

Statement 2  gives us a range and we can assume it's a big number. If the range includes a prime number then this statement is insufficient too.

Can the statement 2 have a prime number?

If  x=15!+2 , then, we can factor out 2 since 15! has 2 in it and there's also +2. Subsequently, we can divide that number by 2.

We can factor 2 and it becomes  2

We can divide this number by 2 and, thus, have a factor  n=2 , which is smaller than the number  x=15!+2 .

This pattern continues throughout the range. Any number we choose for  x  within this range can be divided in such a way that it has a factor  n  smaller than itself (i.e.,  1<n<x ).

This statement always gives us a definite YES answer, unlike statement 1, which sometimes gives a yes and sometimes a no.

Thus, statement 2 is sufficient.

Answer: B

Trap:  This question can overwhelm you with theory and make you lose sight of what's actually needed. That's why it's crucial to stay focused on the main requirement and keep referring back to the question as you work through the problem.

Minhaz03 · Answer

Regarding Statement 2, can 15!+7 or 15!+11 be a prime? How do we understand that it's not possible? Apologies if I could not understand from the previous replies.

Bunuel · Answer

None of those can be prime. For example, 15! + 7 can be factored as 7(2 * 3 * 4 * 5 * 6 * 8 * 9 * 10 * 11 * 12 * 13 * 14 * 15 + 1), which shows that 15! + 7 is not prime, as it is the product of two integers greater than 1.

If x is an integer, does x have a factor n such that 1 < n < x?

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