Try this question out and see how you do.

Which of the following is NOT a factor of ?

- 17
- 21
- 25
- 29
- 42
- 51
- 85
- 121

Factorials can be intimidating. First off, it’s very difficult to calculate factorials over 6, much less find the factors. However, when finding the factors of a factorial, we don’t want to multiply the number up and start slogging away trying to figure out factors.

Instead, for illustrative purposes, let’s write the numbers out and see which combinations of factors you can find. For example, let’s start with a relatively manageable factorial: 8! Let’s actually write this out 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8.

Now let’s say I ask you to find the largest value of n, where is a factor of 8! The way to go about doing this is to find out how many twos there are in 8! Obviously, there is the number 2. But what other 2’s do we have? Four is comprised of two 2s (2 x 2). And the number 8 is comprised of three 2s (). Finally, there is the two from 6, 2 x 3.

Counting the number of 2s in 8! we would get 7. Therefore, is a factor of 8! because, when you divide 8! by ), you are left with an integer.

Now, to return to original we want to see which numbers we can pluck from 20! Our goal is to see which answer choices contain factors that are found in 20! Let’s start with (A) 17. There is definitely one 17 in 20!. With (B) 21, we can see that 21 is made up of 3 x 7, two integers that are found in 20! So far, not too bad. Just make sure not to omit answer choice (B) because it is greater than 20. Again, we are combining the factors in 20! to see which of the answer choices we can make from those factors.

Answer (C) 25 is composed of 5 x 5. Are there two 5s in 20! ? Yes, the integer in 5, and the 5 in 10 (5 x 2).

Answer (D) is the prime number 29. Is there a 29 in the product of the first 20 integers? No. Therefore, (D) is not a factor and is one answer.

Answer (E), 42, is made up of 2 x 21. We know that 21 is already a factor. We have plenty of 2s in 20!. So (E) is definitely a factor.

Answer (F) 51 is tricky. You should know that 51 is a multiple of 17—17 x 3, to be exact.

Answer (G) 85 is also a multiple of 17—17 x 5.

Finally, answer (H) 121, contains . Are there two 11s in 20! ? Well, there is clearly one 11. But in order to get the second eleven, we would have to multiply up to 22 (11 x 2). But 20! stops at 20 so we aren’t able to get that extra eleven. Therefore, (H) is also an answer.

Final Answers: D. and H.

If you understand this problem, take a crack at problem 21 on page 343 in the revised GRE book. It’s similar, but has its own unique twist. Let’s see if you can get it!

And if you want more GRE math practice, check out Magoosh GRE Prep. We have over 400 revised GRE math practice questions, each with a video explanation. Here’s a sample problem on our GRE-like interface: Square Root and Sum

Hello,

I am a bit confused because I thought that the factors of 20 were 1,2,4,5,10, and 20. I asked 2 of my school’s online tutors about this and they couldn’t give me an answer. Thanks, hope to hear from you soon!

Hi Jake,

The twist here is the (!) sign, which means you have to multiply a number times each positive integer less than it. For example:

3! = 3 x 2 x 1

5! = 5 x 4 x 3 x 2 x 1

So 20! is 20 x 19 x 18… x 1

Now I think the problem will make more sense. If not, let me know.

But 20! stops at 20 so we aren’t able to get that extra eleven. Therefore, (H) is also an answer.

I don’t understand how it stands to reason that since we ARE NOT able to get the extra eleven that H is an answer. So, if we could get that extra 11, does it mean that H isn’t an answer? I was kind of following you up until this point, but now I’m more confused than I was before. Off to google and see if I can find a better explanation.

Hi James,

If you read the original question it say which of the following is NOT a factor of 20!. 121 = 11 x 11 is not a factor of 20! because there is only one 11 (1 x 2…. x 10 x 11… x 20). Therefore 121 is NOT a factor so it one of the answers.

Let me know if that’s clear.

Thanks, sir..

Great! I’m happy the problem was helpful.

hi, thank you very much for this cogent explanation, quite helpful!

Great! I’m happy it was helpful. This problem was spun off of an ETS problem from The Official Guide, so you know this type of problem is definitely in their bag of tricks!