# Lesson: Consecutive Integers

## Comment on Consecutive Integers

### Hello

Hello
In below question
N = abc where a, b and c are the hundreds, tens and units digit respectively. If a, b and c are non-zero consecutive numbers such that a < b < c, then which of the following must be true?
I.N is always divisible by 2
II.N is always divisible by 3
III.N is divisible by 6 only if b is odd.

If we apply the logic (n-1)n(n+1) then it does not hold true for above case.

Can you help me understand why not applicable to above case.

Regards,
Abhimanyu ### Be careful. abc does not

Be careful. abc does not represent the product of three numbers.

Each variable (a, b, and c) represents a DIGIT in the 3-digit number abc. As such the rules related to the product (n-1)n(n+1) do not hold.

### Then how it can be solved

Then how it can be solved ### The question: N = abc where a

The question: N = abc where a, b and c are the hundreds, tens and units digit respectively. If a, b and c are non-zero consecutive numbers such that a < b < c, then which of the following must be true?

I. N is always divisible by 2
II. N is always divisible by 3
III. N is divisible by 6 only if b is odd.
------------------------------------------
IMPORTANT:
Property #1) If an integer is divisible by 3, then the sum of its integers will be divisible by 3.

Property #2) If an integer is divisible by 6, then it must be divisible by 3 AND 2.

To get a feel for the properties of these 3 consecutive integers, let's...
Let x = a the 1st number (aka a)
So, x+1 = the 2nd number (aka b)
And x+2 = the 3rd number (aka c)

So, the sum a + b + c = x + (x+1) + (x+2) = 3x + 3 = 3(x + 1)
Aha, so the SUM of any 3 consecutive integers must be divisible by 3 (since 3(x+1) is definitely divisible by 3)

Now let's examine the statements...
I. N is always divisible by 2
This need not be true
For example, one possible value of N is 345, in which case N is NOT divisible by 2

II. N is always divisible by 3
This IS true
We already showed that, since the SUM of any 3 consecutive integers must be divisible by 3, then we know that N is definitely divisible by 3

III. N is divisible by 6 only if b is odd.
If b is odd, then c is EVEN, which means N is divisible by 2
Since N is also divisible by 3, we can conclude that N IS divisible by 6

So statements II and III must be true.

### Will the GMAT always

Will the GMAT always explicitly specify if they're asking for consecutive integers, consecutive EVEN integers, and consecutive ODD integers? I got your first reinforcement prob wrong because I thought a, b, c could also be 1, 3, 5, or 2, 4, 6, hence c-a = 2 cannot always equal 2 ### Yes, on test day, the

Yes, on test day, the question will explicitly state the kinds of consecutive numbers.
Consecutive integers ...-3, -2, -1, 0, 1, 2, 3, ...
Consecutive EVEN integers: ...-6, -4, -2, 0, 2, 4, 6, ...
Consecutive ODD integers: ...-5, -3, -1, 1, 3, 5,...

### Hey Brent,

Hey Brent,

In the question above, I just wanted to confirm that the answer is A because the integers 0*1*2 (E-O-E) is still divisible by 4 as 0 is divisible by 4 (or any number)?

Neel ### Question Link: https:/

You are correct; 0 IS divisible by 4 (in fact, 0 divisible by all integers)

If x and y are integers, we can say that x is divisible by y, if we can write x = yk, where k is some integer.

For example, 12 is divisible by 3, because we can write 12 = (3)(4), and 4 is an integer.
And 70 is divisible by 5, because we can write 70 = (5)(14), and 14 is an integer.
Likewise, 0 is divisible by 0, because we can write 0 = (4)(0), and 0 is an integer.

Does that help?

Cheers,
Brent

### Hi

Hi

Is Zero divisible into any number with no remainder? ### Yes, that's correct. However,

Yes, that's correct. However, I've never seen an official GMAT question that requires this property. In fact, the GMAT test-makers typically restrict the values in Integer Properties question to POSITIVE integers only.

### It is an interesting fact.

It is an interesting fact. Zero is surely an integer full of wonders. Thanks Agreed!

### Hi Brent, would you happen to

Hi Brent, would you happen to have a different explanation for data sufficiency question 317 in OG 18? I’m just not sure why the numbers 6,4, and 2 were chosen ### Question link: https:/

Statement 1: The integers are consecutive odd numbers
Let's examine some consecutive odd numbers: 1, 3, 5, 7, 9, 11, 13....

Notice that each integer is 2 greater than the number before it.

So, if we let x = the first number...
then x + 2 must be the next integer
and x + 2 + 2 must be the integer after that.
And so on.

Does that help?

Cheers,
Brent

Here's my full solution: https://gmatclub.com/forum/the-sum-of-4-different-odd-integers-is-64-wha...

### Makes a lot more sense.

Makes a lot more sense. Thanks!

### Hi Brent, could you please

Hi Brent, could you please address Bunuel's approach for the following question? It seems to contradict the formula you have shared in this video.
(Question: https://gmatclub.com/forum/if-k-is-a-positive-integer-what-is-the-remainder-when-k-2-k-3-k-242852.html)

Bunuel states that: "There is a rule saying that The product of n consecutive integers is always divisible by n!."

However, in this video you tell us that the Rule states that "Every nth # is divisible by n."

From my understanding, being divisible by n! and being divisible by n are two very different operations.

Thank you! ### Bunuel's rule doesn't

Bunuel's rule doesn't contradict the rule in the video lesson. In fact, his law is very similar to mine.

Keep in mind that, within the product of n consecutive integers, we also have the product of n-1 consecutive integers.

For example, consider the product of 5 consecutive integers: (3)(4)(5)(6)(7)
This product must be divisible by 5

Also, within these 5 consecutive integers, we have the product of 4 consecutive integers, (3)(4)(5)(6), which means (3)(4)(5)(6) must be divisible by 4, which means (3)(4)(5)(6)(7) is divisible by 4 as well.

Likewise, within these 5 consecutive integers, we have the product of 3 consecutive integers, (3)(4)(5), which means (3)(4)(5) must be divisible by 3, which means (3)(4)(5)(6)(7) is divisible by 3 as well.

And so on.

So.....

Let's say K = the product of n consecutive integers.
As such, K must be divisible by n

Also, within the n consecutive integers, there are n-1 consecutive integers.
As such, K must be divisible by n-1

Also, within the n consecutive integers, there are n-2 consecutive integers.
As such, K must be divisible by n-2

etc

So, K must be divisible by n, n-1, n-2, n-3, . . . . 2, 1
In other words, K must be divisible by n!

Cheers,
Brent

### Awesome! Thank you for going

Awesome! Thank you for going out of your way to explain with that example. Really appreciate it as it makes perfect sense now.

### Hi Brent, quick question here

Hi Brent, quick question here.

I am reading Bunuel's and Pacifist85 explanations, and both of them use 0 as the first multiple of 10. That is a surprise for me because I used to think multiples of 10 are: 10, 20, 30 and so on. Can then we say that 0 is the first multiple of any number since we can divide 0 by any number?

Thanks! ### 0 is a multiple of all

0 is a multiple of all integers.
That said, pretty much all official GMAT questions involving integer properties will limit the values to positive integers.

### https://gmatclub.com/forum

https://gmatclub.com/forum/what-is-the-sum-of-the-integers-from-100-to-200-inclusive-274500.html#p2122336

Hi Brent,

Can you please show me what process to follow to solve this problem please!

Thanks
Fatima-Zahra 