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## Comment on

Consecutive Integers## 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

## 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

## 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,

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

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)?

Thanks in advance!

Neel

## Question Link: https:/

Question Link: https://gmatclub.com/forum/if-n-is-a-positive-integer-is-n-3-n-divisible...

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

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.

## Agreed!

Agreed!

## Hi Brent, would you happen to

## Question link: https:/

Question link: https://gmatclub.com/forum/the-sum-of-4-different-odd-integers-is-64-wha...

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.

## Hi Brent, could you please

(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

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