A certain class of questions tends to have more going on than might be apparent on the surface. (I’m being intentionally vague as to the certain class “ I’ll tell you what it is after you’ve tried the problem!)

Give yourself approximately 2 minutes to try the below GMATPrep problem. When you’re done, take a look at it again and ask yourself, What was this testing? What was it hiding?

* If

nis a positive integer andris the remainder when (n“ 1)(n+ 1) is divided by 24, what is the value ofr?(1)

nis not divisible by 2.(2)

nis not divisible by 3.

Got something for me? Sure?

La la la. I’m just adding words here so that you don’t inadvertently glance down and see the answer while you’re still figuring things out up above. : ) Okay, what are the clues? *Integer* and *remainder* tell us that this is likely a number properties problem “ this is the class I was referring to earlier. I can tell this is number properties from a couple of key words, but it turns out there’s even more going on. The words *divided* *by* bring up the idea of divisibility. Finally, the problem begins by talking about the variable *n*, but also later mentions *n* “ 1 and *n* + 1. Put those three terms together and what have we got? Consecutive integers!

So we’re going to need to think about consecutive integer properties for 3 numbers in a row, and yet the divisibility info in the question stem talks only about the first and third numbers, while the info in the statements refers to the middle number. Okay.

Are any rules popping up in your mind right now? What have you learned about consecutive integers in the past, in particular for a set of 3 consecutive integers?

Let’s see. The product of any 3 consecutive integers is divisible by 2, because at least one of the numbers in the set has to be even. Likewise, the product has to be divisible by 3, because at least one of the numbers in the set must be a multiple of 3.

Finally, before we start playing with the statements, think about that 24. What do we need to know about it or do with it? 24 = 2 Ã— 2 Ã— 2 Ã— 3, so the actual math to be done is:

If the stuff on top contains three 2′s and a 3, then the bottom will cancel out entirely and the remainder will be zero. If something on the bottom can’t cancel out, then there will be some kind of remainder greater than zero. In other words, it would be useful if we were able to figure anything out about the factors of *n*, *n* “ 1, and *n* + 1.

Okay. Anything else? Ready to tackle the statements?

(1)

nis not divisible by 2.

We could follow two different approaches here: testing numbers or using theory. Either way, if *n* is an integer but is not divisible by 2, then *n* has to be odd. Let’s try the testing numbers method first.

We need an odd number. Let’s say *n* = 3. Then (*n* “ 1)(*n* + 1) = (2)(4) = 8. Okay, 8 / 24 = 0 remainder 8. If *n* = 5, then (*n* “ 1)(*n* + 1) = (4)(6) = 24. Okay, 24 / 24 = 1 remainder 0. Different remainders this statement is not sufficient. Eliminate answers A and D.

(2)

nis not divisible by 3.

We’ll continue with our testing numbers approach.

Let’s see *n* is not divisible by 3. Let’s say *n* = 2. Then (*n* “ 1)(*n* + 1) = (1)(3) = 3. Okay, 3 / 24 = 0 remainder 3. Let’s say n = 5 “ hey, we already did this one when testing statement 1. We know the answer gives a remainder of 0. Once again, two different remainders means this statement is not sufficient. Eliminate B.

(1)

nis not divisible by 2.(2)

nis not divisible by 3.

Here’s where things can get a little messy. We need to pick something odd and not divisible by 3. Let’s see *n* = 5 fits these criteria and we already know that has a remainder of 0. Great. What’s another possibility? *n* = 7 could work. Then (*n* “ 1)(*n* + 1) = (6)(8) = 48. Okay, 48 / 24 = 2 remainder 0. We just got another remainder of 0. Does that mean the remainder will always be zero? Or does that mean we just haven’t tried the right number yet?

This is the one place where the testing numbers approach breaks down a bit. When we keep getting the same answer, there’s no way to know for sure whether we’ve found a consistent answer or whether we just haven’t tried the right numbers yet the only way to know for sure is to use theory.

In the moment, with the clock ticking, try one more example and, if you get zero yet again, assume you’ve found a pattern and choose your answer accordingly. When the test is over, though, figure it out for sure.

Here’s the theory:

Let’s go back to statement 1 by itself. If *n* is odd, then the sequence of three consecutive integers must be even, odd, even.

Interesting. If *n* “ 1 is even and so is *n* + 1, then the product of the two must contain two 2′s. It’s divisible by 4.

Further, those two even numbers are consecutive. As a result, one of the two must actually be a multiple of 4, not just 2. (Think about why this has to be true. Write out some real numbers to test it.) In other words, (*n* “ 1)(*n* + 1) is divisible not just by two 2′s but by three 2′s. It’s actually divisible by 8!

Important: memorize this. From now on, just know that the product of two consecutive even integers must be divisible by 8. Then you can skip the time it took to figure that out.

Statement 1 isn’t sufficient by itself because, although we know we can divide out those three 2′s on the bottom, we have no idea what will happen with the 3. Why?

Exactly one of the three terms *n* “ 1, *n* , or *n* + 1 must be divisible by 3. If the divisible by 3 number turns out to be *n* “ 1 or *n* + 1, then their product will by divisible by the necessary three 2′s and a 3, leaving a remainder of 0. If, on the other hand, the divisible by 3 number turns out to be *n*, then the product of (*n* “ 1)(*n* + 1) will not be divisible by 3, leaving a remainder greater than 0.

If you’re going for a really high quant score, memorize this too. If you know that two consecutive even integers will definitely be divisible by 8 but may or may not be divisible by 3, then you can assess this statement in about 20 seconds.

What about statement 2? If *n* is not divisible by 3, then it could be even or odd. Either way, one of (*n* “ 1) and (*n* + 1) must be divisible by 3 (remember one of our basic rules about 3 consecutive integers: one of them must be divisible by 3). So we can definitely divide out the 3. What about the three 2′s?

If *n* is even, then (*n* “ 1) and (*n* + 1) will be odd, as will their product. That odd product won’t divide out any 2′s at all, leaving a remainder greater than 0.

If, on the other hand, *n* is odd and (*n* “ 1) and (*n* + 1) are even, then the product will divide out all of the 2′s (we just proved this rule earlier). We also know the 3 will be divided out, so this product would leave a remainder of 0. We’ve got conflicting remainders, so this statement is not sufficient by itself.

Slap the two statements together. When evaluating statement 2, we had two scenarios: one in which *n* was odd and one in which *n* was even. Statement 1 tells us that n is odd, so we can drop the even scenario in statement 2. That leaves us with the odd scenario: (*n* “ 1) and (*n* + 1) are even, so their product is divisible by 8, and one of the two is also divisible by 3 (since *n* is not). In other words, we can always divide out all three 2′s and the 3, leaving a remainder of 0. Together, the two statements are sufficient.

The correct answer is C.

**Important Note**: Don’t use theory if you haven’t already figured out and memorized those extra consecutive integer rules ahead of time. Use real numbers and just take the risk that the pattern you find when combining the two statements will last.

If you do learn the full theory in advance, though, then you can use theory to do this problem very quickly. It’ll boil down to this:

Statement 1: *n* is odd. (*n* “ 1) and (*n* + 1) are even, so they’re divisible by 8. The product (*n* “ 1)(*n* + 1) may or may not be divisible by 3, which will lead to a remainder of either 0 or something else “ not sufficient.

Statement 2: *n* is not divisible by 3. It might be even or odd. One of (*n* “ 1) and (*n* + 1) is divisible by 3, then. We can’t tell whether the product of (*n* “ 1)(*n* + 1) is odd (in which case, it’s not divisible by any 2′s, leaving a remainder greater than 0) or even (in which case, it is divisible by three 2′s, leaving a remainder of 0). Not sufficient.

Statements 1 and 2: *n* is odd, so the product (*n* “ 1)(*n* + 1) is divisible by 8. *n* is not divisible by 3, so the product (*n* “ 1)(*n* + 1) is divisible by 3. Therefore, the product (*n* “ 1)(*n* + 1) will always cancel out all three 2′s and the 3, so the remainder will always be 0.

**Key Takeaways for Using Theory vs. Testing Numbers**

(1) Testing numbers is often easier but has one significant drawback: what to do when we keep getting the same answer for each case we test. In that circumstance, without using theory, we can’t be 100% certain whether the statement actually is sufficient or whether we just haven’t tested the right number or numbers to get a contradictory answer.

(2) In the moment, while the clock is ticking, try one more set of numbers. Actively try to pick numbers that are weird / different to increase the odds that you’ll find the contradictory answer if it exists. If you still get the same answer, then assume the statement is sufficient and move on. Afterwards, though, go back and try to prove it using theory.

(3) When you do prove something via theory, memorize it! (That is, if it’s not too hard to memorize – otherwise, take the time to memorize only if you’re going for a very high score.) In this case, we now know these rules about consecutive integers:

- The product of three consecutive integers is divisible by 2 and by 3.

- The product of two consecutive even integers is divisible by 8.

- The product of two consecutive even integers is divisible by 8 and by 3 if the odd number between them is not divisible by 3.

* GMATPrep questions courtesy of the Graduate Management Admissions Council. Usage of this question does not imply endorsement by GMAC.

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“That leaves us with the odd scenario: (n â€“ 1) and (n + 1) are even, so their product is divisible by 8, and one of the two is also divisible by 3 (since n is not). In other words, we can always divide out all three 2â€™s and the 3, leaving a remainder of 0.”

Stacie, Can you please detail this statement : “one of the two is also divisible bu 3 (since n is not).” I am confused with that statement.

Sure!

Pick any set of three consecutive integers that you want. Are any divisible by 3? How many? Pick another set of three consecutive integers and repeat the exercise. Do it again. Notice any patterns?

In any set of 3 consecutive integers, by definition, exactly one of them is divisible by 3. 1, 2, 3. 2, 3, 4. 3, 4, 5. Even -1, 0, 1 (because zero is divisible by 3!).

So, in our question above, if n itself is not divisible by 3, then either the integer before it or the one after it has to be divisible by 3.

I love this blog post and this problem. Thanks Stacey!

I think theory is way better than testing cases for this particular question – but only if you understand the theory deeply.