# Lesson: Divisor Rules

## Comment on Divisor Rules

### Hello Brent,

Hello Brent,
In the link below

https://gmatclub.com/forum/is-a-multiple-of-139044.html
You wrote " Given: If the average (arithmetic mean) the 4 numbers is 30 "
How did u come to this inference from Question stem
Regards,
Abhimanyu

### Good catch. When I answer

Good catch. When I answer Data Sufficiency questions, I paste in a pre-formatted template, and that template has that information about the average. I forgot to delete that part of my template.

I have since edited my response. Thanks!

### Hello Brent,

Hello Brent,
Need one clarification.
When we say if x is a divisor of m and n both means m/x, n/x then ( m+n)/x or ( m-n)/x.

Then in the below example why it can't be true
that when (2k-4) is divisible by 7 then 2k/7 and 4/7 is possible.

Regards,
Abhimanyu

### We need to be careful here.

We need to be careful here. This is an IF-THEN construction.

So, for example, IF x is a divisor of m and n, THEN x is a divisor of m+n, and x is a divisor of m-n. For example, since 5 is a divisor of 25 and 15, we know that 5 is a divisor of 25+15 and 25-15.

We can't necessary reverse that order though.
For example, even though 3 is a divisor of 5+4, we can't conclude that 3 is a divisor of 5 and 3 is a divisor of 4.

### Hello Brent, thanks for your

Hello Brent, thanks for your exposure so far, can you kindly shed more light on the above rule: for instance, "if 5 is a divisor of 35, then 5 is a divisor of (35)(47), please how did you get 47, in the first example, 98 in the second example and the third example. Please can you provide some insights?

### Hi fobembe,

Hi fobembe,

Those values (47, 98, etc) are arbitrary numbers; they could be ANY integers.

For example, if 5 is a divisor of 35, then 5 must be a divisor of ANY product of 35 and some other integer. So, 5 must be a divisor of the following products:
(35)(11)
(35)(6587)
(35)(-29)
(35)(789,006)
etc.

Likewise, since 13 is a divisor of 39, then we know that 13 must be a divisor of ANY product of 39 and some other integer. So, 13 must be a divisor of the following products:
(39)(8)
(39)(734)
(39)(18,211)
(39)(-41)
etc.

Does that help?

### Can you someone help me with

Can you someone help me with this? I can't understand this.

If x is an integer and y = 3x + 2, which of the following CANNOT be a divisor of y?

A. 4
B. 5
C. 6
D. 7
E. 8

### APPROACH #1

APPROACH #1
Test various possible values of x and y

If x = 1, then y = 5.
So, 5 CAN be a divisor of y
ELIMINATE B

If x = 2, then y = 8.
So, 4 CAN be a divisor of y, and 8 CAN be a divisor of y
ELIMINATE A and E

If x = 3, then y = 11.

If x = 4, then y = 14.
So, 7 CAN be a divisor of y
ELIMINATE D

By the process of elimination, the correct answer is C

------------------------------------

APPROACH #2
For a number to be divisible by 6, the number must by divisible by 2 AND by 3

The question tells us that y = 3x + 2
In other words, y is TWO GREATER than some multiple of 3
In other words, y is NOT divisible by 3
If y is NOT divisible by 3, then y is NOT divisible by 6

### If x is an integer and y=3x+2

If x is an integer and y=3x+2, which of the following CANNOT be a divisor of y?

A) 4
B) 5
C) 6
D) 7
E) 8

if X = 1 Y = 5
In my opinion 4 (A) is not a divisor of 5, hence the correct answer is C??

Could you maybe explain what I am missing here?

### Yes, the correct answer is C.

Yes, the correct answer is C.

When x = 1, y = 5
Since 5 IS a divisor of 5, we can ELIMINATE B

When x = 2, y = 8
Since 4 and 8 are both divisors of 8, we can ELIMINATE A and E

Keep going to get C as the final answer.

Here's my full solution: https://gmatclub.com/forum/if-x-is-an-integer-and-y-3x-2-which-of-the-fo...

### Hi Brent, failing to

Hi Brent, failing to understand what the question is asking for? https://gmatclub.com/forum/for-any-integer-n-greater-than-1-n-denotes-the-product-of-168575.html

### This is a Strange Operator

This is a Strange Operator question, which is covered here: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

GIVEN: [n denotes the product of all the integers from 1 to n, inclusive.
So the "[" symbol tells us to do something to the value next to it.

For example: [6 = the product of all the integers from 1 to 6, inclusive.
That is, [6 = (1)(2)(3)(4)(5)(6)

Likewise: [8 = the product of all the integers from 1 to 8, inclusive.
That is, [8 = (1)(2)(3)(4)(5)(6)(7)(8)

The question asks "How many prime numbers are there between [6 + 2 and [6 + 6, inclusive?"
In other words, "How many prime numbers are there between (1)(2)(3)(4)(5)(6) + 2 and (1)(2)(3)(4)(5)(6) + 6?"

Does that help?

Cheers,
Brent

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

https://gmatclub.com/forum/a-b-and-c-are-positive-integers-if-a-b-and-c-are-assembled-into-218938.html

### Hi Brent, does the reverse

Hi Brent, does the reverse hold true for all the rules in this lesson? eg: if k is a divisor of N+M, is k a divisor of both N and M? Thanks

### No, the reverse does not hold

No, the reverse does not hold true in this case.
For example, 5 is a divisor of (7 + 3), but 5 is NOT a divisor of either 7 or 3.

Cheers,
Brent

### Hi Brent,

Hi Brent,

One of the questions linked from gmatclub goes like this:
Is x – y divisible by 5?

(1) x is divisible by 5.
(2) y is divisible by 5

OA: C

In case of x=y we get a result of 0 to the equation x - y. Why can we say that 0 is divisible by 5?

Thanks,
Beni

### Question link: https:/

Great question!

In general, we can say that N is divisible by d if there exists an integer k such that N = dk

For example, we know that 15 is divisible by 5, because there exists an integer k such that 15 = 5k
In fact that integer is k = 3
In other words, since we can write: 15 = (5)(3), we know that 15 is divisible by 5.

Likewise, we know that 0 is divisible by 5, because there exists an integer k such that 0 = 5k
In fact that integer is k = 0
In other words, since we can write: 0 = (5)(0), we know that 0 is divisible by 5.

In fact, we can say that 0 is divisible by all positive integers.

Having said all of that, it is VERY unlikely that the GMAT would ever test this concept. In fact, the test-makers typically restrict all values to POSITIVE integers when it comes to divisibility questions.

Cheers,
Brent

### Hi Brent,

Hi Brent,

If an integer n is to be selected at random from the integers 1 to 100, inclusive, what is the probability that n(n + 1) will be divisible by 4 ?

A. 1/4
B. 1/3
C. 1/2
D. 2/3
E. 3/4

### Your video begins by making

Your video begins by making the assumption that all divisor variables are integers. Does the GMAT give questions in which the divisors are non-integers or could be assumed to be a non integer?

### The GMAT will specify whether

The GMAT will specify whether the values are integers. So, if there is no mention of a variable being an integer, then we cannot assume that it's an integer.

That said, for pretty much all integer properties questions, the test-makers will restrict all values to positive integers.

Cheers,
Brent

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

https://gmatclub.com/forum/if-n-20-17-then-n-is-divisible-by-which-of-the-143638.html

I'm not too sure how to approach this one, would this be an appropriate approach?

20! + 17 could be rewritten as (20x19x18x17)(16! + 1)

As such it is clear that this is divisible by 20,19,18 and 17. I'm not sure what is divisible within that bracket but I don't think that's a line the GMAT crosses.

Is this approach logical?

### Question link: https:/

We can't write 20! + 17 as (20x19x18x17)(16! + 1)

When we expand the expression, we get: 20! + 20x19x18x17, which is NOT the same as 20! + 17

Here's my full solution: https://gmatclub.com/forum/if-n-20-17-then-n-is-divisible-by-which-of-th...

Cheers,
Brent

### I’m having a tough time with

I’m having a tough time with this question- checked your answer on gmat club but still need a little help. I’m looking at only the answer choice 15 to try to understand the methodology first. I understand the “some number” concept but why is it some number PLUS 1? Does that matter in this question?

### Solution link: https:/

This all comes down to factoring.
To set things up consider these examples of factoring:
15x + 15 = 15(x + 1)
(15)(21) + 30 = 15(21 + 2)
(15)(14)(13) + 15 = 15[(14)(13) + 1]
(15)(14)(13)(12)(11) + 15 = 15[(14)(13)(12)(11) + 1]
(15)(14)(13)(12)(11)...(3)(2)(1) + 15 = 15[(14)(13)(12)(11)...(3)(2)(1) + 1]

Noticed that if we add a 2 to the left hand side, we're not obligated to include it as part of the factorization.
We can just write: (15)(14)(13)(12)(11)...(3)(2)(1) + 15 + 2 = 15[(14)(13)(12)(11)...(3)(2)(1) + 1] + 2

Does that help?

### Yes completely. Actually your

Yes completely. Actually your response was so obvious now I feel silly! Thank you!

### Not to worry; it happens to

Not to worry; it happens to EVERYONE (I'm no exception :-)

### Hi Brent,

Hi Brent,

Can you please check if my approach is valid? (I am more interested in my logic with statement 1) & 2) combined)

Statement 1) since we are not told anything about x being an integer or not, we can say that if 5x is divisible by 11, then 5x must be a multiple of 11 ==> the smallest number x can take is 11:5=2.2. Now obviously we have at least one case where when x=2.2, x is not divisible by 11, if x=11 then x is divisible by 11. Hence we cannot answer the question with certainty.
Statement 2) applying same logic we would not get a definitive yes or no to the question.

Statement 1) & 2) combined. Am I understanding it is correct that when we have two equations in data sufficiency questions, one equation per statement, then we can add them up? Here is my logic: we know that 7x=11n, and 5x=11m, if 7x is equal to 11n, then we can safely assume that 7x-11n=0, same for 5x, 5x-11m=0, since we know that both conditions must be met we can add two equations together and we will get ==> 12x-11n-11m=0 => 12x-11(n+m)=0. We can clearly conclude that 12x must be equal to 11(n+m), we know that n and m are integers and since we know that 12x must be equal to 11*some integer, for this to hold water 12x must be multiple of 11, and 11y (where y=n+m) must be multiple of 12 (2*2*3*x = 11*y), left side must have at least one 11, and right side must have at least two 2s and one 3. Hence we can say that x will always be a multiple of 11, which means it will always be divisible by 11. Is my solution good?

### Question link https:/

Perfect logic - nice work!!

Cheers,
Brent

### Hey Brent,

Hey Brent,

I would appreciate your input. These 2 questions are confusing me:

https://gmatclub.com/forum/is-the-integer-b-divisible-by-219192.html

https://gmatclub.com/forum/is-x-11-an-integer-278593.html

The first one is very clear. In regards to the second one: In the case of 5x/11, which is like 5 times x/11: Since 5 is an integer, x/11 has to be an integer as well? Isn´t there a rule or something? I really fell into that and didn´t get it right. I saw your explanation in the post and that makes a lot of sense, still I am wondering if I am confusing rules here?

Cheers,

Philipp

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

Your question: "Since 5 is an integer, x/11 has to be an integer as well?"
Not necessarily.
(5)(1.2) = 6
So, if x/11 = 1.2, then (5)(x/11) will be an integer.

I think you're thinking of a couple of other properties regarding integers.
For example, if x + 3 = some integer, then x must be an integer.
Likewise, if x - 7 = some integer, then x must be an integer.

Does that help?

Cheers,
Brent

### Thanks, I guessnI just

Thanks, I guessnI just confused it with some other rules.

### https://gmatclub.com/forum/30

https://gmatclub.com/forum/30-20-20-20-is-divisible-by-all-of-the-following-267144.html

Is there any typo when you tested option C? 30^20 = (10^20)(4^20).

### Solution link: https:/

Good catch! I've edited my response.
Thanks for the heads up!!