Most people feel that the topic of number properties is hard or at least a little tricky. The reason is that no matter how much effort you put into it, you will still come across new concepts every time you sit with some 700+ level problems of this topic. There will be some concepts you don’t know and will need to “figure out” during the actual test. I came across one such question the other day. It brought forth a concept I hadn’t thought about before so I decided to share it today:

Say you have N consecutive integers (starting from any integer). What can you say about their sum? What can you say about their product?

**Say N = 3**

The numbers are 5, 6, 7 (any three consecutive numbers)

Their sum is 5 + 6 + 7 = 18

Their product is 5*6*7 = 210

Note that both the sum and the product are divisible by 3 (i.e. N).

**Say N = 5**

The numbers are 2, 3, 4, 5, 6 (any five consecutive numbers)

Their sum is 2 + 3 + 4 + 5 + 6 = 20

Their product is 2*3*4*5*6 = 720

Again, note that both the sum and the product are divisible by 5 (i.e. N)

**Say N = 4**

The numbers are 3, 4, 5, 6 (any five consecutive numbers)

Their sum is 3 + 4 + 5 + 6 = 18

Their product is 3*4*5*6 = 360

Now note that the sum is not divisible by 4, but the product is divisible by 4.

**If N is odd then the sum of N consecutive integers is divisible by N, but this is not so if N is even.**

Why is this so? Let’s try to generalize – if we have N consecutive numbers, they will be written in the form:

(Multiple of N),

(Multiple of N) +1,

(Multiple of N) + 2,

… ,

(Multiple of N) + (N-2),

(Multiple of N) + (N-1)

In our examples above, when N = 3, the numbers we picked were 5, 6, 7. They would be written in the form:

(Multiple of 3) + 2 = 5

(Multiple of 3) = 6

(Multiple of 3) + 1 = 7

In our examples above, when N = 4, the numbers we picked were 3, 4, 5, 6. They would be written in the form:

(Multiple of 4) + 3 = 3

(Multiple of 4) = 4

(Multiple of 4) + 1 = 5

(Multiple of 4) + 2 = 6

etc.

What happens in case of odd integers? We have a multiple of N and an even number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will always add up in pairs to give the sum of N:

1 + (N – 1) = N

2 + (N – 2) = N

3 + (N – 3) = N

…

So when you add up all the integers, you will get a multiple of N.

What happens in case of even integers? You have a multiple of N and an odd number of other integers. The other integers are 1, 2, 3, … (N-2) and (N-1) more than a multiple of N.

Note that these extras will add up to give integers of N but one will be leftover:

1 + (N – 1) = N

2 + (N – 2) = N

3 + (N – 3) = N

…

The middle number will not have a pair to add up with to give N. So when you add up all the integers, the sum will not be a multiple of N.

For example, let’s reconsider the previous example in which we had four consecutive integers:

(Multiple of 4) = 4

(Multiple of 4) + 1 = 5

(Multiple of 4) + 2 = 6

(Multiple of 4) + 3 = 3

1 and 3 add up to give 4 but we still have a 2 extra. So the sum of four consecutive integers will not be a multiple of 4.

Let’s now consider the product of N consecutive integers.

In any N consecutive integers, there will be a multiple of N. Hence, the product will always be a multiple of N.

Now take a quick look at the GMAT question that brought this concept into focus:

*Which of the following must be true?*

*1) The sum of N consecutive integers is always divisible by N.*

*2) If N is even then the sum of N consecutive integers is divisible by N.*

*3) If N is odd then the sum of N consecutive integers is divisible by N.*

*4) The Product of K consecutive integers is divisible by K.*

*5) The product of K consecutive integers is divisible by K!*

*(A) 1, 4, 5*

*(B) 3, 4, 5*

*(C) 4 and 5*

*(D) 1, 2, 3, 4*

*(E) only 4*

Let’s start with the first three statements this question gives us. We can see that out of Statements 1, 2 and 3, only Statement 3 will be true for all acceptable values of N. Therefore, all the answer choices that include Statements 1 and 2 are out, i.e. options A and D are out. The answer choices that don’t have Statement 3 are also out, i.e. options C and E are out. This leaves us with only answer choice B, and therefore, B is our answer.

This question is a direct application of what we learned above so it doesn’t add much value to our learning as such, but it does have an interesting point. By establishing that B is the answer, we are saying that Statement 5 must be true.

*5) The product of K consecutive integers is divisible by K!*

We will leave it to you to try to prove this!

(For more advanced number properties on the GMAT, check out Parts **I**, **II**, **III**, **IV** and **V** of this series.)

*Getting ready to take the GMAT? We have free online GMAT seminars running all the time. And, be sure to follow us on Facebook, YouTube, Google+, and Twitter!*

*Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the **GMAT** for Veritas Prep and regularly participates in content development projects such as this blog!*

The product of K consecutive integers is divisible by K!

I tried following 3 cases and seemed to hold:

1) 99*100*101/(3*2) = 33*50*101

2) 99*100*101*102/(4*3*2) = 33*25*101*51

3) 36*37*38*39*40/(5*4*3*2) = 12*37*38*39

Thus I selected that this property holds. :-)