Happy Thursday! Here’s a look back at Monday’s GRE Brain Twister!

13! – 11! is divisible by n but not by n^2. What is the range of values of n if n is a prime number?

(A) 20

(B) 24

(C) 26

(D) 29

(E) 31

## Explanation

The first step is to factor this large number: 11!(13×12 – 1) = 11! (155) = 11!(31)(5). We know that n has to be a prime. Be careful though, because n^2 cannot be a factor of the number. Notice how 5, which is a prime, occurs twice in 11! (the number 5 and in the number 10: 5 x 2). The same holds true for the lowest primes, 2 and 3. There are plenty of 2s that can be factored from 11! and a few 3s as well. 7, however, only occurs once in 11!. Therefore, it is the lowest number that fits the criteria and the number 31 that we factored from 155 at the beginning is the highest prime. This gives us 31-7 = 24, answer (B).

## Answer: B

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Hello! I want to ask if the question itself means that we have to find the total number of primes that are within the range which satisfies problem? If that’s so, is 31-7+1= 25 an accepted answer?

Hi Sally,

This question is asking for the range of numbers, not the quantity of prime numbers. In fact, if we were to ask how many prime numbers satisfy these conditions, we would have to find out how many prime numbers are between 5 and 31. The primes that are between 5 and 31 are: 5, 7, 11, 13, 17, 23, 29, 31. This is a total of 9 🙂

Correction: The least prime number we found was 7 (not 5) and the correct range of prime numbers between 7 and 31 would be: [7, 11,13, 17, 19, 23, 29, 31], a total of 8.

If the question is asking for a range of numbers that would satisfy n, then our answer should be 8.

Hi Vatwani,

Yes, you are correct! This was our mistake 🙂

5 would NOT be included in this range, so there would be a total of 8 prime numbers that satisfy this question.

Thanks for checking this question and providing us this feedback–students like you keep up on our toes and help improve Magoosh as much as possible 🙂

How did you get the first step in the factorization: 11!(13×12 – 1)

Hi Jackson,

In this case, we need to find the prime factorization of (13*12-1), which is 155. We need to break ‘155’ down into only prime numbers. To find the prime factorization, we can start to test the divisibility of 155 by each prime number starting from ‘2’. 155 is not divisible by 2, so we move on to 3. Since it’s not divisible by 3, we move on to 5. 155 is divisibly by 5, and that gives us 155=5*31. Since 31 is also a prime number, we can’t break this down any more: 5*31 is the prime factorization of 155. I hope that makes sense! For more information on this method and how to do it, see this blog post: https://magoosh.com/gre/2011/gre-math-essential-tips-for-factoring/ 🙂

” Therefore, it is the lowest number that fits the criteria and the number 31 that we factored from 155 at the beginning is the highest prime. This gives us 31-7 = 24, answer (B).”

I don’t understand this. Why are we finding the difference between the highest prime and n which is 7 in this case. Could you please explain.

It sounds like you want to know why subtracting 7 form 21 gets you the range for n in this problem. I’ll be happy to explain.

First, recall the rules for the possible values of n: For a prime number to belong to the set n, 13!-11! must be divisible by n, and 13!-11! must NOT be divisible by the n squared.

To find all prime numbers that can go evenly into 13!-11!, we first need to prime factorize 13!-11!. This, as you saw, gets us 11!(31)(5). So we see that 31 is be the highest prime number in the set of numbers divisible by 13!-11!, and 5 is the lowest prime.

Now, if an additional 31 or 5 are hidden inside 11!, then those extra numbers can be used to make 31^2 or 5^2, which would go evenly into 11!(31)(5). This would mean that 31 and 5 aren’t n values. You can already clearly see that 31 in not in 11!, because 31 is greater than 11. But what about 5?

Well, let’s look at 11!, turned into a string of multiplied numbers: 11*10*9*8*7*6*5*4*3*2*1. You can see that there is a 5 inside 11! There are two fives in fact, because the 10 in 11! equals 5*2. So 5 won’t work for n.

To find the real lowest value for n, check the other prime numbers in 11! 3 and 2 both occur more than once, because the 6 in 11! is equal to 3*2. This leaves us only with 7 and 11. Each of these prime numbers occurs once in the the 11!(31)(5) set, so 11 and 7 can both be n. 7, the lower of these two values, must then be the lowest value for n.

So 7 is the lowest n value, and 31 is the highest. And the range for n equals (highest n) – (lowest n). In this case, that is 31-7, or 24.

Range of values should be the actual prime values which are the factors of (13!-11!). As mentioned 31-7=24 doesn’t include all the values as prime values. I didn’t understand the concept of range of prime number.

Btw, It’s really brain twister 😀

The set of values that satisfy the given criteria are {7,11, 31}. 31 being the highest and 7 being the lowest. Therefore the range of such values(concept we learn in statistics) is 31 – 7 = 24.