# Lesson: Divisor Rules

## Comment on Divisor Rules

### Hello Brent,

Hello Brent,

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 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? 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? 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? 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). 