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## Comment on

Divisor Rules## 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,

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

## 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

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

Answer: C

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

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

## 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

please explain

## Here's my solution: https:/

Here's my solution: https://gmatclub.com/forum/a-b-and-c-are-positive-integers-if-a-b-and-c-...

Cheers,

Brent

## Hi Brent, does the reverse

## 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,

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:/

Question link: https://gmatclub.com/forum/is-x-y-divisible-by-240256.html

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,

Need your help:

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

## Here's my solution: https:/

Here's my solution: https://gmatclub.com/forum/if-an-integer-n-is-to-be-selected-at-random-f...

Cheers,

Brent

## Your video begins by making

## 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

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:/

Question link: https://gmatclub.com/forum/if-n-20-17-then-n-is-divisible-by-which-of-th...

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

## Solution link: https:/

Solution link: https://gmatclub.com/forum/if-n-20-17-then-n-is-divisible-by-which-of-th...

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

## Not to worry; it happens to

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

## Hi Brent,

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

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:/

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

Perfect logic - nice work!!

Cheers,

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

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

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

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

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

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

## Solution link: https:/

Solution link: https://gmatclub.com/forum/30-20-20-20-is-divisible-by-all-of-the-follow...

Good catch! I've edited my response.

Thanks for the heads up!!