# 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 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 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,

Thanks
Fatima-Zahra ### Hey Brent,

Hey Brent,

is there a rule that allows us to conclude that (k+1)(k)(2k+1) is DIVISIBLE by 4?

Regarding this Q:

https://gmatclub.com/forum/if-n-is-a-positive-integer-is-n-3-n-divisible-by-109941.html

And generally, is a product of 3 consecut. numbers also always divisible by 6?

Thanks,

Philipp There's no rule that says (k+1)(k)(2k+1) is DIVISIBLE by 4
Notice that, if k = 2, then (k+1)(k)(2k+1) = (3)(2)(5) = 30, and 30 isn't divisible by 4.

There is a rule that says the product of n consecutive integers MUST be divisible by n, n-1, n-2,..., 2 and 1
For example, the product of 6 consecutive integers MUST be divisible by 6, 5, 4, 3, 2 and 1

Likewise, the product of 3 consecutive integers MUST be divisible by 3, 2 and 1
If the product is divisible by 3 and 2, the product is also divisible by 6.

Cheers,
Brent

### So in the question the answer

So in the question the answer is YES to divisibility by 4, since we know that two expressions are even, thus the least possible value being 2,3,4? so Whenever to even numbers are part of the expression, it is divisible by 4? ### Yes, if there are two even

Yes, if there are two even numbers as part of a product, then that product must be divisible by 4.

### Is it considered a

Is it considered a consecutive group of integers if n is an integer and the pattern is as follows: n, n-1,

I mean in descending order

Or when in any GMAT questions the word "consecutive" appears we then must consider it as an incremental trend? ### If we CAN take a set of

If we CAN take a set of integers and arrange those integers in ascending order so that each integer is one greater than the integer before it, then we have a set of consecutive integers.

So, for example {4, 7, 6, 5} is a set of consecutive integers.

I mention this because I've seen questions that state x, y and z are consecutive integers. In these cases, we can't immediately assume that x < y < z.
It COULD be the case that x = 4, y = 2 and z = 3.

Cheers,
Brebt

Thank you