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# GMAT Tip: Breaking Down a GMATPrep Consecutive Integer Problem

Today’s GMAT tip comes from test prep firm ManhattanGMAT. In this article, they share helpful tips on how to solve consecutive integer problems.  Read on to see what they have to say!

This week, we’re 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 n2 – 1 is divided by 8, what is the value of r?

“(1) n is odd.

“(2) n is not divisible by 8.”

The first thing you’ll probably notice: I didn’t include the answer choices. The five Data Sufficiency answer choices are always the same, so we should have those memorized. If you don’t 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 we’ve got that out of the way, let’s tackle this problem! This one’s a theory question; they’re asking us about the concept of consecutive integers (as opposed to asking us to do more straightforward calculations with consecutive integers) and they’re not even nice enough to tell us straight out that this is about consecutive integers! We have to figure that out or – even better – recognize it.

I have two variables, n and r, and I’m 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 can’t find any value or I can find more than 1 value, then the information is not sufficient.

First, I’ll be ahead of the game if I’m able to recognize that n2 – 1 is just another way of saying (n+1)(n-1). The problem also talks about the integer n; put those together and we’ve got (n-1), n, and (n+1), or three consecutive integers. Add n2 – 1 to your list of “Unscramble the Code” rephrasings; seeing this one should make you think “this might be about consecutive integers!,” especially if integer n is also floating around in the problem.

When you’re studying a problem and see something like n2 – 1, and you don’t 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 you’ll 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. Let’s 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. I’m 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 it’s divisible by 4.

But the problem asks us about 8, so I guess this isn’t enough information. Maybe I’ll 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.

What’s my next possibility? 2, 3, 4. The product of 2 and 4 is 8 and 8/8 = 0 with a remainder of 0. For this particular case, r=0 again. That’s interesting.

What’s 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? I’ve 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 here’s 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 2s as factors, not just two 2s. 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, I’m 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, n2 – 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.

What’s the next thing I can try? 1, 2, 3. (Note: make sure to use whatever’s 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… 3 remainder 5. r = 5.

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

Answer to bonus question: No, n2 – 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.

Finally, you’ve 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 don’t want a hint about how to do the problem, then don’t read the takeaways below till after you’ve 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 n2 – 1 is probably the most common indicator and n3 – n is another very common indicator because n3n = n(n2 – 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 you’re 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 you’ll go a lot further if you study how to recognize it in the first place.

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