# What is this Quant Question Hiding?

*by*, Mar 28, 2013

A certain class of questions tends to have more going on than might be apparent on the surface. (Im being intentionally vague as to the certain class Ill tell you what it is after youve tried the problem!)

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

* Ifnis a positive integer andris the remainder when (n1)(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. Im just adding words here so that you dont inadvertently glance down and see the answer while youre 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 theres 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 were 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?

Lets 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 2 2 3, so the actual math to be done is:

[pmath]{(n+1)(n-1)}/{2*2*2*3}[/pmath]

If the stuff on top contains three 2s and a 3, then the bottom will cancel out entirely and the remainder will be zero. If something on the bottom cant 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. Lets try the testing numbers method first.

We need an odd number. Lets 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.

Well continue with our testing numbers approach.

Lets see *n* is not divisible by 3. Lets say *n* = 2. Then (*n* 1)(*n* + 1) = (1)(3) = 3. Okay, 3 / 24 = 0 remainder 3. Lets 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.

Heres where things can get a little messy. We need to pick something odd and not divisible by 3. Lets see *n* = 5 fits these criteria and we already know that has a remainder of 0. Great. Whats 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 havent 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, theres no way to know for sure whether weve found a consistent answer or whether we just havent 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 youve found a pattern and choose your answer accordingly. When the test is over, though, figure it out for sure.

Here's the theory:

Lets 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 2s. Its 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 2s but by three 2s. Its 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 isnt sufficient by itself because, although we know we can divide out those three 2s 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 2s 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 youre 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 2s?

If *n* is even, then (*n* 1) and (*n* + 1) will be odd, as will their product. That odd product wont divide out any 2s 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 2s (we just proved this rule earlier). We also know the 3 will be divided out, so this product would leave a remainder of 0. Weve 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 2s and the 3, leaving a remainder of 0. Together, the two statements are sufficient.

The correct answer is C.

**Important Note**: Dont use theory if you havent 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. Itll boil down to this:

Statement 1: *n* is odd. (*n* 1) and (*n* + 1) are even, so theyre 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 cant tell whether the product of (*n* 1)(*n* + 1) is odd (in which case, its not divisible by any 2s, leaving a remainder greater than 0) or even (in which case, it is divisible by three 2s, 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 2s 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 cant be 100% certain whether the statement actually is sufficient or whether we just havent 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 youll 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.

###### Recent Articles

###### Archive

- February 2020
- January 2020
- December 2019
- November 2019
- October 2019
- September 2019
- August 2019
- July 2019
- June 2019
- May 2019
- April 2019
- March 2019
- February 2019
- January 2019
- December 2018
- November 2018
- October 2018
- September 2018
- August 2018
- July 2018
- June 2018
- May 2018
- April 2018
- March 2018
- February 2018
- January 2018
- December 2017
- November 2017
- October 2017
- September 2017
- August 2017
- July 2017
- June 2017
- May 2017
- April 2017
- March 2017
- February 2017
- January 2017
- December 2016
- November 2016
- October 2016
- September 2016
- August 2016
- July 2016
- June 2016
- May 2016
- April 2016
- March 2016
- February 2016
- January 2016
- December 2015
- November 2015
- October 2015
- September 2015
- August 2015
- July 2015
- June 2015
- May 2015
- April 2015
- March 2015
- February 2015
- January 2015
- December 2014
- November 2014
- October 2014
- September 2014
- August 2014
- July 2014
- June 2014
- May 2014
- April 2014
- March 2014
- February 2014
- January 2014
- December 2013
- November 2013
- October 2013
- September 2013
- August 2013
- July 2013
- June 2013
- May 2013
- April 2013
- March 2013
- February 2013
- January 2013
- December 2012
- November 2012
- October 2012
- September 2012
- August 2012
- July 2012
- June 2012
- May 2012
- April 2012
- March 2012
- February 2012
- January 2012
- December 2011
- November 2011
- October 2011
- September 2011
- August 2011
- July 2011
- June 2011
- May 2011
- April 2011
- March 2011
- February 2011
- January 2011
- December 2010
- November 2010
- October 2010
- September 2010
- August 2010
- July 2010
- June 2010
- May 2010
- April 2010
- March 2010
- February 2010
- January 2010
- December 2009
- November 2009
- October 2009
- September 2009
- August 2009