Decoding the Prime Disguise

How can the GMAT disguise a prime number (or any other) problem? I asked this question a couple of years ago at the start of a very important article entitled Disguising – and Decoding – Quant Problems. Go read that article right now, if you haven’t already. I’ll wait.

Towards the end of that article, I referenced two Official Guide ® problems. I was very excited today to see that one of these problems is part of the free practice problem set that now comes with the new GMATPrep 2.0 ® software – so I can actually reproduce it here and we can try it out!

Disclaimer: this is a seriously challenging problem. Set your timer for 2 minutes, but practice your “1 minute timing” here. If you don’t have a pretty good idea of what’s going on by the halfway mark, try to figure out how to make a guess. Pick an answer by the 2 minute mark (all right, I’ll let you go to 2 minutes 30 seconds if necessary – but that’s all!).

Does the integer k have a factor p such that 1 < p < k?

(1) k > 4!

(2) 13! + 2 ≤ k ≤ 13! + 13”

Did the question stem look completely unfamiliar to you? If yes, did you take my advice and read the earlier article first? If not, go read that article and then try the problem again.

This problem is an excellent example of the importance of figuring out how to decode the GMAT before we ever get into the testing center. A higher-level, well-prepped quant student can look at that sentence and say, “Oh, I know what this is about because I’ve seen something like it before. This is a prime problem.”

How do we know this? The question stem mentions a factor, so we know we have some kind of divisibility problem in general. Divisibility often, but not always, addresses prime as well. Then we’re given the inequality 1 < p < k. They’re telling us that this factor p would have to lie between 1 and k. What kinds of numbers have factors between 1 and themselves?

If you’re not sure, test some real numbers.

All positive integers can be split into three categories:

• the number 1 (exactly one factor)
• prime numbers (exactly two factors, 1 and itself)
• composite numbers (three or more factors: 1, itself, and at least one more factor in between 1 and itself

Composite numbers, by definition, have factors between 1 and themselves. Also by definition, a prime number will never have a factor p that falls in between 1 and itself.

Back to our question:

Does the integer k have a factor p such that 1 < p < k?

It’s really just asking us “is k composite?” because the characteristic described is the third category: is there a factor between 1 and itself?

If we can recognize* that within the first 20 seconds or so, then we might have a shot at answering this question in 2 minutes. If we have to figure this out, then we are definitely not going to finish this thing on time (and maybe we’ll never get there).

*Note: recognize simply means “Oh, I’ve seen something similar to this before, and last time, it meant that the problem was about prime vs. composite, so let me see if that applies here, too… yes, yes, it does! Great. Okay, what do I do with that?”

The question now is “is k composite?” Here’s statement 1:

(1) k > 4!

That exclamation point means to take the factorial of the number 4, so let’s do that: 4×3×2×1 = 24. Statement 1 is simply saying k > 24.

Does that tell us whether k is composite? Nope. There are prime numbers larger than 24 (such as 29) and also composite numbers larger than 24 (such as 25). This statement is insufficient; cross off answers A and D.

Phew. At least we got rid of two answer choices. Unfortunately, now we have to deal with that nasty statement 2.

(2) 13! + 2 ≤ k ≤ 13! + 13

What in the world should we do with that? If I didn’t know the problem was really about prime vs. composite, I’d have absolutely no idea what to do next. Even knowing what the problem is about, I’m still feeling pretty intimidated by this statement.

But wait… I’m having a memory. I’ve seen one of these “prime / composite” problems before and it also had a statement that consisted of a short range of numbers, like this one. (See #39 from the Quant Review 2^nd Edition. Statement 2 in our current problem is the same kind of information as statement 1 in #39: 31 < p < 37.)

The key is that we have a finite number of possible values for k. Last time (on problem #39), I was able to test the numbers in the given range. This time, though, the numbers are way too big. Can we think of any way to test these numbers for composite vs. prime status?

There’s no easy test that will let us tell whether a really large number is prime. But there is a test I can use to tell that a number is definitely composite: I just have to find one factor that is larger than 1 and smaller than itself. For example, any even number greater than 2 is composite, because every even number greater than 2 has 2 as a factor.

Okay, the first number in our range is 13! + 2. Write it out:

13×12×11×10×9×8×7×6×5×4×3×2×1 + 2

Yuck. But I know the 13! number by itself is an even number (because it has 2 as a factor). If I add 2 to an even number, I’m going to get another even number. So this crazy number is divisible by 2, in which case it must be composite. For the first number, at least, I can answer the question: is k composite? Yes, it is.

Thinking through this is leading to a brainstorm. When I was studying primes and divisibility, I remember learning a rather obscure rule, one that I almost never use. It goes like this:

If two numbers are added together and those two numbers share a factor, then their sum shares that same factor.

For example: 428 + 2,986 = ? I don’t particularly want to add that up, but I can see that both numbers have 2 as a factor. Therefore, their sum also has 2 as a factor and so the sum is a composite number.

Here’s another example: 369 + 3,936 = ? I can see that both numbers have 3 as a factor; therefore, the sum of the two numbers also has 3 as a factor and is a composite number.

Interesting! Let’s try it with the next number in the range: 13! + 3. (Unclear where that came from? What’s the first number in the range 2 < x < 6? 2, of course. What’s the next number? 3. How did you get to that next number? It’s just one larger than the previous number. Adding one more each time allows us to get to 4, 5, and 6.)

So, our first number was 13! + 2. Our next number is 13! + 2 + 1, or 13! + 3.

13×12×11×10×9×8×7×6×5×4×3×2×1 + 3

Check it out. 3 is a factor of the first crazy number and it’s also a factor of 3. So, whatever that crazy sum is, 3 is a factor. Therefore, the sum is definitely composite!

What about the next number?

13×12×11×10×9×8×7×6×5×4×3×2×1 + 4

4 is a factor of both numbers again, so this is another composite number. Hmm, I think I’m starting to see the pattern. We keep adding 1: +5, +6, etc. all the way up to +13. (If you don’t see it, yet, keep writing out each subsequent number and examining the results.)

The number 13! includes every integer from 1 to 13, and our “extra added” numbers run from 2 to 13. In other words, every single number in the range is going to fit this pattern: the two numbers share a factor and so we can deduce that their sum shares that same factor.

In other words, every number in this range is a composite number. We have a definitive “yes” answer! Statement 2 is sufficient!

The correct answer is B.

Take-Aways

1) If you’re going for a very high quant score, you’ve got to learn to recognize the things you see; you’re not going to have time to figure everything out from scratch in 2 minutes. Decoding a problem thoroughly can take 5, 10, even 20 minutes sometimes!

2) Know the definition of prime, but also know the difference between prime and non-prime (or composite) numbers. Further, know how the test writers can ask about prime vs. composite without actually using either word.

3) Don’t stop with this concept. What other concepts can they ask about without using the specific word or words we usually expect? And how can we learn to recognize those in future, too?