Sum/Product of Consecutive Integers | Number Properties

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 $\times$ 6 $\times$ 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 $\times$ 3 $\times$ 4 $\times$ 5 $\times$ 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 $\times$ 4 $\times$ 5 $\times$ 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

(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!

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