• Take advantage of our featured Black Friday/Cyber Monday deals

  • Take advantage of our featured Black Friday/Cyber Monday deals


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!

If n is a positive integer and r is the remainder when [pmath]n^2 [/pmath] 1is divided by 8, what is the value of r?

(1) n is odd.

(2) n is 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) n is 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) n is 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 12th 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.