To begin, here are four reasonably challenging practice problems.
1) S is a set of n consecutive positive integers. Is the mean of the set a positive integer?
Statement #1: the range of S is an even integer
Statement #2: the median of S is a positive integer
2) If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?

 (A) 315

 (B) 330

 (C) 345

 (D) 360
 (E) 375
I. 4
II. 6
III. 18

 (A) I only

 (B) II only

 (C) I and II only

 (D) II and III only
 (E) I, II, and III

 (A) 5

 (B) 10

 (C) 25

 (D) 30
 (E) 35
Explanations for these will come at the end of the article.
Consecutive Integers
The word “consecutive” means “in a row; one after the other.” A set of three consecutive integers might mean {3, 4, 5} or {137, 138, 139} or {–25, –24, –23}. As long as the integers are in a row, it doesn’t matter whether they are big or small, positive or negative. In fact, the set {–1, 0, +1} contains one positive number and one negative number (of course, zero is neither positive nor negative).
Properties of Consecutive Integers
a) Any set of n consecutive integers will contain exactly one number divisible by n. For example, any three integers in row must contain a multiple of 3; any 17 integers will contain one multiple of 17, etc. Now, you may look at the set {–1, 0, +1}, a set of three consecutive integers, and wonder: where is the multiple of 3? This is tricky. As it turns out, zero is a multiple of every integer, because (any integer) times zero equals zero.
b) In a set of three consecutive integers, we could have two evens and one odd, or two odds and one even, depending upon where we started. In a set of four consecutive integers, we would have to have two evens and two odds. More generally, if we have an odd number of consecutive integers, we could have more evens or more odds, depending on the starting value, but if we have an even number of consecutive integers, the evens and odds have to be evenly split.
c) If n is an odd number, then the sum of n consecutive integers is divisible by n. For example: for any three integers in a row, the sum is divisible by 3; for any 7 integers in a row, the sum is divisible by 7, etc.
Algebraic Representations
Anyone can recognize that {6, 7, 8, 9, 10, 11} is a set of consecutive integers. When you are given plain old numbers, it’s easy to see whether they are consecutive integers. That’s too easy. The GMAT will not ask about that. Instead, the GMAT will give you algebraic representations of consecutive integers.
The following are examples of algebraic representations of consecutive integers:
{n, n + 1, n + 2, n + 3, n + 4, n + 5}
{n – 2, n – 1, n, n + 1}
{n + 12, n + 13, n + 14}
For all of those, if n equals any integer, then the set will be a set of consecutive integers. For simplicity, let’s pick n = 10. The first set becomes the set of integers from 10 to 15; the second, from 8 to 11; and the third, from 22 to 24. Even without knowing the value of n, we can apply the properties of consecutive integers to the set: for example, the second is a set of four consecutive integers, so it must have two evens and two odds; the third is a set of three consecutive integers, so the sum of those three numbers must be divisible by three.
Summary
If you had some insights reading this article, you may want to give the problems above a second look before reading the solutions below. Here’s another practice question from inside Magoosh:
5) http://gmat.magoosh.com/questions/943
If you have questions about anything unclear in this post, please let us know in the comments section.
Practice problem explanations
1) First of all, for a set of consecutive integers, or for any set of evenly spaced numbers, the mean and the median are equal. If there’s an odd number of members of the list, then the median is the middle number. If there’s an even number of members of the list, then the median is the average of the two middle numbers. For example, the median of {1, 2, 3, 4, 5} is 3, a positive integer and member of the set. For consecutive integers, an even number of members would mean that the mean or median is the average of the two middle integers. For example, the median of {1,2, 3, 4} is the average of 2 and 3, that is, 2.5, not an integer. The only way the mean or median can be an integer is if the set of consecutive integers has an odd number of members.
Statement #1: If there are an even number of consecutive integers, then the evens and odds are balanced in the set, and the first and last number must be opposite: one must be even and the other must be odd. Thus, the range, the difference of (max) – (min) would be either (even) – (odd) or (odd) – (even), in either case, an odd number. If the range is odd, the number of consecutive integers is even.
If there are an odd number of consecutive integers, then the first and last numbers are either both even or both odd. The range would be either (even) – (even) or (odd) – (odd), in either case, an even number. If the range is even, the number of consecutive integers is odd. That must be the case here. As we have seen above, this means the mean or median is a positive integer. This statement, alone and by itself, is sufficient.
Statement #2: As we discussed above, the mean = the median. If the latter is a positive integer, so is the former. This statement, alone and by itself, is sufficient.
Both statement are separately sufficient. Answer = (D)
2) When we have a set of consecutive integers or consecutive multiples of the number, the range depends only on the size of the set, how many members, not where on the number line the set starts or ends. For example, any seven consecutive integers will have a range of 6, whether it’s 1 through 7 or 51 through 57. Thus, we can ignore the starting number, 255, which is just there to confuse us. We can pick any more convenient starting value.
Let’s start at a1 = 15 = 15*1. Then a2 = 15*2 = 30, and a3 = 15*3 = 45. Continuing in this pattern, the last number would be a23 = 15*23. Don’t multiply that yet. The range would be highest minus the lowest:
range = (a23) – (a1) = 15*23 – 15*1 = 15*(23 – 1) = 15*22
Now, use the doubling & halving trick. Half of 22 is 11, and twice 15 is 30, so
15*22 = 11*30 = 330
Answer = (B)
3) This is a tricky one. If you start plugging in values for n, you are sunk. The numbers are gigantic and unwieldy. This one begins with some clever factoring. Clearly, in the first factor, we can factor out a factor of n:
In the second one, we can us the Difference of Two Squares:
Now, put all that together, and rearrange the order:
Written in that order, we see this is a product of four consecutive integers. We absolutely know that two of the numbers are even and two are odd. If there are two even numbers, each one has a factor of two, so the product would have a factor of 2*2 = 4. The product must be divisible by 4. We know I must be true.
Any three consecutive integers contains one multiple of 3, so four consecutive integers would contain at least one. We have a multiple of 3, which has 3 as a factor, and at least one even number, which has 2 as a factor. They may be the same number or may not be: that doesn’t matter. As long as the entire product contains at least one factor of 3 and at least one factor of 2, then the product must be divisible by 6. We know II must be true.
To be divisible by 18=2*3*3, we need one factor of 2 and two factors of 3. We definitely have more than one factor of 2. The problem is: how many factors of three do we have? In a set of four consecutive integers, we could have two factors of three, as in {12, 13, 14, 15} or {6, 7, 8, 9}. BUT, we could also have a set of four consecutive integers with only one factor of three, as in {4, 5, 6, 7}. Thus, we could have two factors of three, and the product could be divisible by 18, but we are not sure. We cannot say that this must be true. Therefore, III cannot be among the answer.
Only I and II work. Answer = (C)
4) We know that odd integers are spaced two apart: if we have one odd integer, we can add 2 to get the next one. Starting from the given expression for the first member of the set, we can say:
This highest member, we can set equal to 7n
n = 2 or n = 5. Well, if n equals 2, then the first member would be 4, not an odd integer. That value doesn’t work with the problem. Therefore, we know n = 5. This means that first member is 25, and the whole set is {25, 27, 29, 31, 33, 35}. The median of this set is the average of the two middle numbers, 29 and 31; median = 30.
Answer = (D).
Hey. Not really a question. Great explanations. Really helped a lot. I noticed in question 3’s explanation, when talking about 18’s divisibility, you gave an example of sets that contained 2 numbers having factors of 3 and you mentioned {7,8,9,10}. Dunno if you meant {6,7,8,9} instead. Made me scratch my head abit. I’m just pointing this out to be corrected if it was a typo. Thanks 👍🏼
Hi Chinenye!
You are absolutely right–we meant {6,7,8,9}! I’ve fixed the typo, and I’m sorry for the confusion it has caused. Thanks so much for pointing this error out for us so that we could fix it for other students. 😀
Cheers!
What is the reason that sum of consecutive integers will not be divisible by n if its an even number?
Hi Apoorva,
Since you’re a Premium student, I forwarded your message on to our team of student support specialists. They will be able to provide you with an answer shortly! Just so you know, you’ll receive the answer in an email 🙂
is this why when adding up any series of consecutive positive numbers the intermediate values can never be a prime number … for example when adding 1 to 100 … the intermediate values (except for the first one at times) will never be a prime …
3 6 10 15 21 28 36 45 55 66 78 91 105 120 136 153 171 190 210 231 253 276 300 325 351
and so forth … 99 intermediate values in all … the final one being 5,050 … none are prime numbers (except the first one) …
Basically, yes. When n is even and more than two, the sum of the consecutive integers is also even, making the sum nonprime if it’s bigger than 2. And if n is odd, the sum of the consecutive integers will be divisible by n, making that sum not prime, unless n is 1 and the single number is 1, since a prime numbers are divisible by 1. (In that case, though, you technically wouldn’t have a sequence.) So in essence, once a sequence gets rolling, there are no real chances to have a prime sum. Only the first sum in a sequence can sometimes be prime.
“Now, from our consecutive integers rules, we know that if we multiply a set of consecutive integers together, the product will be divisible by the number of units in that set, and every integer less than that until 1. So, for example, in a set of four consecutive integers, we know for certain the number will be divisible by 4, 3, and 2.”
The above is stated in the explanation section from a problem in the Consecutive Integers video, however, this rule is not stated in the video or anywhere else. The rules stated in the video are:
1. A set of n consecutive integers will always contain one number divisible by n.
2. If n is odd, then the sum of n consecutive integers will always be divisible by n.
3. In a set of 4 consecutive integers, two integers must be even and two integers must be odd.
Is there a fourth rule?
Bracha,
This could be treated as a fourth rule, yes. And in fact, there are a number of other patterns that apply to consecutive integers that aren’t stated in the video lesson in particular. The Magoosh GRE Premium video lesson (also available in Magoosh GMAT Premium) was meant to give an overview of the most basic, foundational rules. This additional rule you’ve mentioned from the answer explanation is important too, of course. 🙂
median of 5 consecutive numbers is N. find mean number (please help me with this question)
Hi Muhammed,
If we have a set of 5 consecutive numbers, then they are evenly spaced around the median, meaning that the median and the mean are equivalent. So the mean would also be N. 🙂
Is 1 a multiple of 2 consecutive odd interger ( 1 and 1 ) ?
Please reply.
Because i m really confused about positivity and negativity of multiples.
Thankyou 🙂
1 would be a multiple of two consecutive odd integers, yes. As Mike mentions, it doesn’t matter where the integers fall on the number line. Consecutive is consecutive, whether the consecutive numbers are all negative, all positive, or cross the “zero barrier” between negative and positive.
Question about #3
I expanded the formula (n^2 2n)(n^2 1) = n^4 – n^2 2n^3 + 2n
Because each part has n, I can factor out n and am left with: n(n^3 – 2n^2 – n + 2)
Since n is a factor, I assumed this equation would be divisible by any number, so therefore the answer would be E (I, II, and III)
Where did I go wrong in my math?
Hi Allen,
Happy to help 🙂 You expanded the expression correctly and also correctly factored out n from all of the terms, leaving you with
n(n^3 – 2n^2 – n + 2)
Now, the thing is is that this expression will always be divisible by n. However, that doesn’t mean than the expression will be divisible by any number. In the problem, n can be any integer greater than 50. If n = 51, for example, the expression
51(51^3 – 2*51^2 – 51 + 2)
will be divisible by 51. That doesn’t guarantee that it will also be divisible by any number. For a value of n, the expression must have all of the prime factors of a given number for the expression to be divisible by that number. While many numbers would include all of the prime factors of 18 (2, 3, 3), this is not true for all numbers greater than 50, such as 51.
Does this make sense? I hope this clears up your doubts, but if not, please let us know 🙂
http://gmat.magoosh.com/questions/943
i think the answer to this question is wrong! the question was consecutive integers not consecutive odd integers, so why stating that w=1, x=3, y= 5?
If you try placing those numbers in yw=2 that is 51= 4 not 2…
Hi Liz,
Happy to help! 🙂 In that problem, the part where we test w=1, x=3, y=5 is specifically to prove the same point you’re arguing from a different angle. We are testing an alternative set of numbers with an arithmetic mean of x to see if Statement 2 is sufficient. It turns out that there are multiple sets of numbers with a mean of x, but these do not guarantee that w, x, and y are consecutive numbers. This is why we determine that statement 2 is insufficient.
I hope that helps! 🙂
I don’t think the answer is correct for Question 1.
If the set has an number of terms, it does not mean the mean and median will definitely be even. For example, the set can be {1, 2, 3} with 2 as the mean and median; while set {2, 3, 4} will have 3 has the mean and median. Therefore, statement 1 is insufficient.
Am I missing anything here?
Hi Chi Le,
First, as Mike explains, for a set of consecutive integers (or for any set of evenly spaced numbers) the mean and the median are equal, not even. As you showed in your examples, if there’s an odd number of members of the list, then the middle number is the mean and the median of the set 🙂
Now, if the range of the set is odd, then the number of consecutive integers is even. The range of both sets of numbers you mentioned is even:
{1, 2, 3}: range = 31 = 2
{2, 3, 4}: range = 42 = 2
Let’s contrast this by looking at a set with an even number of consecutive integers:
{1, 2, 3, 4}: range = 41 = 3
As we can see, the range is odd. This reflects the general property that the range is odd for a set with an even number of consecutive integers.
Statement (1) states: The range of S is an even integer.
Since the range is even, we can conclude that there is an odd number of integers in the set, so that mean (and median) will be an integer. Since the entire set itself is composed of positive integers, then we know that the mean and median will be a positive integer. Therefore, Statement (1) is sufficient to answer the question “Is the mean of the set a positive integer?”
Hope this clears up your doubts!
Hi, I don’t understand why the starting number does not matter for consecutive integers/multiples.
ex. {2,4,6} –>range=4
{4*2, 4*4,4*6} –>range=16
While both sets have same consecutive multiples, the range is different as the starting points are different.
Hi there 🙂
I can understand this can be confusing. You’re right that the ranges of the two sets you’ve listed are different. With that in mind, the idea isn’t that the ranges are the same (they’re not!) but rather that we can determine the range knowing that the set is composed of consecutive multiples, regardless of what the multiple is. To do this, we need to know how many numbers are in the set and the multiple we’re considering. We can then factor out that multiple to see that we are left with a series of consecutive integers multiplied by the multiple. So, the range will be the difference of the last and first number in the set of consecutive integers multiplied by the multiple under consideration. Let’s look at the sets you listed to see what I mean 🙂
{2,4,6} (multiples of 2)
Factor out 2 (the multiple): 2{1,2,3} <–this is the product of the multiple and the set of 3 consecutive integers
Range: (31)*2 = 4 or 62 = 4
{8, 16, 24} (multiples of 8)
Factor out 8 (the multiple): 8{1,2,3} <–again we have the product of the multiple and the set of 3 consecutive integers
Range: (31)*8 = 16 or 248 = 16
In both cases, we were able to determine the range by considering the number of integers and multiplying the difference of the last and first by the multiple under consideration. For that reason, the starting number doesn't "matter"–rather it is the number of integers in the set and the multiple that determine the range.
I hope this helps 🙂
Hello Mike,
I have a question on the extra problem linked in this article (http://gmat.magoosh.com/questions/943):
If w, x and y are integers such that w < x < y, are w, x and y consecutive integers?
(1) y – w = 2
(2) The average (arithmetic mean) of w, x and y is x
The OA is A but I am confused given the following:
(1) states yw=2, however the problem does not say anything about the integers being positive or negative, this statement could be satisfied with y = 2 and w = 5 (given that W<Y) and they would NOT be consecutive integers.
I would really appreciate your feedback.
Greetings,
Iván.
Nevermind, I had an awful mistake. Sorry about that!
Mike,
Question 4..
I used your logic stated earlier that since the list is an even number list (6 consecutive odd integers) then the range was odd and the median had to be the average of the two middle values; which should be an even number. And since the the lowest value is n squared then the integers must be positive values.
This logic would void 5, 25, and 35.
So between 10 and 30 left over, selecting 10 would be too low because setting n to 2 or 3 puts 10 too high on a list of 6 consecutive odd integer list to be the median.
So the answer must be 30.
Don’t know if this logic will work every time or if it is sound; any thoughts?
Thank you,
Garrett
Garrett,
That’s Great! That’s a logic I hadn’t considered when I wrote the questions, but it’s perfectly valid. It may not always be applicable, and in better written questions, it will not narrow down so much, but certainly it will always be mathematical correct and may yield some insights.
Mike 🙂
Thank you Mike. I have a couple of questions in regard to the explanation for Q# 1:
1) you mentioned that “the median of {1,2, 3, 4} is the average of 3 and 4, that is, 3.5, not an integer.” Shouldn’t the median, instead, be the average of 2 and 3, that is, 2.5?
2) It’s stated that “the only way the mean or median can be an integer is if the set of consecutive integers has an odd number of members.” But what about the consecutive set {2,4,6,8}, where the mean and median is 4+6/2 = 5? Here we have an even number of members and still got a integer for mean/median?
Best,
Dear Mohammed,
Great questions. As for #1, it was a typo, and I just corrected it. Good eye. As for #2, I am discussing sets of consecutive integers, a very specific kind of consecutive set. If we have consecutive evens (as you do) or consecutive odds, or consecutive multiples of something, then the rules are different. That statement is clearly and explicitly about consecutive integers only.
Does this make sense?
Mike 🙂
Awesome! Thank you, Mike 🙂
Dear Mohammed,
You are quite welcome! 🙂 Best of luck to you!
Mike 🙂
With regards to the 3rd practice problem … i think ,the greater than 50 clause has been intentionally inserted to prevented to prevent students from plugging in numbers…infact as soon as we can resolve the function into 4 factors,we can put in any number there..because the product of four consecutive positive integers will always obey the same rules..greater than 50 or not..so if we can just plug in a number like 3..we can get the answer in a few seconds as the calculations will be very simple..
Dear Arjun,
You are absolutely correct. 🙂 The “greater than 50” clause is not actually pertinent to the deep mathematical pattern occurring here, and an astute student, such as yourself, will recognize that. Wellwritten questions discriminate between folks who truly know the math and folks who rely on superficial tricks, such as simply plugging in whenever variables appear. Here, if you see the pattern, you don’t have to plug in at all, so that constraint is irrelevant. Good job at understanding this problem!
Mike 🙂
Mike,
A slightly different take on problem 3. We can use the property that n! must be a factor of the product of ‘n’ consecutive integers. So, in the problem above, the product of four consecutive integers must be divisible by 24 and its factors – 1,2,3,4,6,12. Hence 18 gets eliminated.
Thanks,
Sriram.
Dear Sriram,
YES! That’s quite true, and that’s a very efficient way of approaching that problem! Thank you for sharing this.
Mike 🙂
Q#1 whr the options?????but u stated answer D………..plz mention the options…….
Dear Sanjoy,
I’m happy to respond. 🙂 That question type, with “Statement #1” and “Statement #2” is a Data Sufficiency question. Almost half the math questions on the GMAT Quant section will be DS. It’s very important to be familiar with this type of question, to recognize it instantly, and to have the answer choices memorized. You can find the answer choices and read more about this question type here:
https://magoosh.com/gmat/2013/gmatdatasufficiencytips/
Does this make sense?
Mike 🙂
thnx ….gr8……myself is a GRE prep student……in GRE, am i gonna face this kinda data sufficiency?????what i know there is a data analysis(gre) typo….
Dear Sanjoy,
Ah, I understand! 🙂 On the GRE Quant section, you will face four question formats: ordinary Multiple Choice, Numerical Entry, Multiple Answer, and Quantitative Comparisons; but you will NOT see Data Sufficiency, which is a format unique to the GMAT. Data Interpretation is a category of content on the GRE Quant section, which can appear in different question formats. As you may know, GMAT math is a notch harder than GRE math, so looking at GMAT math is a good way to prepare yourself for the GRE. Just be advised that some of the math you see here will be in Data Sufficiency form: even though you will not need to handle this question on the GRE, it might be worthwhile to learn a little about it just to sharpen you mathematical & logical skills. Some of the logical skills of GMAT DS are similar to skills you might use in GRE QC: for example, sometimes it’s not necessary to do an exact calculation in order to answer the question. Incidentally, some of the GMAT Integrated Reasoning material would give you good preparation for the GRE Data Interpretation questions.
Does all this make sense?
Mike 🙂
gr8 gr8 advice thnx……
Dear Sanjoy,
You are quite welcome. Best of luck to you, my friend.
Mike 🙂