# Breaking Down a GMATPrep Consecutive Integer Problem

*by*, Jan 1, 2011

This week, were going to talk about what to know for consecutive integer problems, and how to recognize what to do on future problems of the same type.

This one is from GMATPrep. Set your timer for 2 minutes. and GO!

Ifnis a positive integer andris the remainder when [pmath]n^2 [/pmath] 1is divided by 8, what is the value ofr?(1)

nis odd.(2)

nis not divisible by 8.

The first thing youll probably notice: I didnt include the answer choices. The five Data Sufficiency answer choices are always the same, so we should have those memorized. If you dont have them memorized yet, add this to your to do list.

Just in case, here are the five choices (in casual language, not official language):

(A) statement 1 works but statement 2 does not work(B) statement 2 works but statement 1 does not work

(C) the statements do NOT work alone, but they DO work together

(D) each statement works by itself

(E) nothing works, not even using them together

Okay, now that weve got that out of the way, lets tackle this problem!

This ones a theory question; theyre asking us about the *concept* of consecutive integers (as opposed to asking us to do more straightforward calculations with consecutive integers). And theyre not even nice enough to tell us that this is about consecutive integers! We have to figure that out or, even better, recognize it.

In this case, we have two variables, *n* and *r,* and are being asked to find the value of *r*. My task is to determine whether I can find one definitive value for *r*. If I can, the information is sufficient. If I cant find any value, or I can find more than 1 value, then the information is not sufficient.

First, Ill be ahead of the game if Im able to recognize that [pmath]n^2 [/pmath] 1 is just another way of saying (*n*+1)(*n*-1). The problem also talks about the integer *n*; put those together and weve got (*n*-1), *n*, and (*n*+1), or three consecutive integers. Add [pmath]n^2[/pmath] 1 to your list of Unscramble the Code rephrasings; seeing this one should make you think this might be about consecutive integers! Especially if the integer *n* is also floating around in the problem.

When youre studying a problem and see something like [pmath]n^2[/pmath] 1, and you dont recognize any special significance to that term, go back over the problem again carefully and figure it out. Use your books and online resources to help you but figure out the significance of that thing. What concept are they *really* talking about here? Next time, maybe youll actually just be able to recognize it right when you see it.

Okay, so now we know that something is going on with consecutive integers here. Lets figure out what it is. Also, keep in mind that the smallest possible value for *n* is 1 (because the question stem tells us that *n* is a positive integer).

The question is asking us to take the first and third consecutive integers and multiply them together, then divide by 8 and see what the remainder, *r*, is.

(1)nis odd.

Hmm. If n is odd, what else can I figure out? Well, *n*-1 and *n*+1 must both be even. Im supposed to multiply those two even numbers together, so whatever the result is, it must be divisible by 2. Actually, it must also be divisible by 4, because each of the two numbers is separately divisible by 2. In other words, the product (*n*+1)(*n*-1) has at least two factors of 2 so its divisible by 4.

But the problem asks us about 8, so I guess this isnt enough information. Maybe Ill just explore things a little bit further, though

So, we could have 0, 1, 2 as our three consecutive integers. In that case the product of 0 and 2 is 0. 0/8 = 0 with a remainder of 0. For this particular case, *r*=0.

Whats my next possibility? 2, 3, 4. The product of 2 and 4 is 8 and 8/8 = 1 with a remainder of 0. For this particular case, *r*=0 again. Thats interesting.

Whats my next possibility? 4, 5, 6. The product of 4 and 6 is 24 and 24/8 = 3 with a remainder of 0. For this particular case, *r*=0 yet again! All right. Is this a pattern or did I just get lucky that the first three all gave the same remainder?

What have I been doing? Ive been taking two consecutive even integers and multiplying them together, then dividing by 8. The two starting integers are each divisible by 2, as I figured out before oh, and heres the new thing! Every *other* even integer, by definition, is a multiple of 4, not just a multiple of 2. 0, 2, 4, 6, 8, 10, 12 starting with zero, every other one is a multiple of 4. So, whenever I multiply two consecutive even integers, one of them has to be a multiple of 4.

What does that mean for the product? The product has to contain three 2's as factors, not just two 2's. The product will gain at least one 2 from the plain even integer and the product will gain at least two 2s from the multiple of 4 even integer. That means the product must be a multiple of 8! So, whenever I divide by 8, Im going to have a remainder of zero. Bingo! Statement 1 is sufficient. Eliminate answers B, C, and E.

I have another thing to add to my Unscramble the Code rephrasing list: When n is odd, [pmath]n^2[/pmath] 1 must be divisible by 8. (Bonus question: is this also true when *n* is even? Try it out and check the end of the article for the answer.) Next time I see something like this, I can just recognize it!

Moving on to statement 2:

(2)nis not divisible by 8.

Hmm. The middle integer of my three consecutive integers is *not* divisible by 8. What are some possibilities?

I can re-use my first possibility from statement 1: 0, 1, 2. When we tested this case before, the remainder, *r*, was zero.

Whats the next thing I can try? 1, 2, 3. (Note: make sure to use whatevers possible next, NOT simply what we used for the previous statement. The previous statement had different parameters.) So, the product of 1 and 3 is 3, and 3/8 is... 0 remainder 3.*r*= 3.

Great! I have (at least) two different possible values for *r*, so I know this statement is insufficient. Eliminate answer D.

The correct answer is A.

Finally, youve got homework! Set your timer for two minutes again, open up your Official Guide 12^{th} Edition, and do problem #170 in the data sufficiency section. What are the similarities? What are the differences? In what ways, specifically, is problem #170 harder than the one we just tried above? (Spoiler alert: *If you dont want a hint about how to do the problem, then dont read the takeaways below till after youve tried it!)*

## Key Takeaways for Solving Consecutive Integer Problems:

(1) They can test you on a concept without naming that concept. Be able to recognize a consecutive integer problem in disguise. The form [pmath]n^2[/pmath] 1 is probably the most common indicator and [pmath]n^3[/pmath] n is another very common indicator because [pmath]n^3[/pmath] *n* = *n*([pmath]n^2[/pmath] 1 ) = *n*(*n*+1)(*n*-1).

(2) Also be able to recognize some of the common useful pieces of information that might be given on consecutive integer problems. Knowing something as seemingly simple as whether certain terms are even or odd can make a big difference, especially if the question deals with divisibility or remainders.

(3) If you do the figuring out work while youre studying, you will save yourself a ton of time and mental effort on the test. You may be able to figure out a decent proportion of this stuff during the test, but youll go a lot further if you study how to recognize it in the first place.

**Answer to bonus question**: No, [pmath]n^2[/pmath] 1 does not have to be divisible by 8 when *n* is even. In fact, it will never be divisible by 8. If *n* is even, then both *n*+1 and *n*-1 are odd. Odd numbers never have 2 as factors, and we need three 2s in order for something to be divisible by 8.

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

###### Recent Articles

###### Archive

- July 2020
- June 2020
- May 2020
- April 2020
- March 2020
- 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