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If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
12 Aug 2018, 16:08

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Question Stats:

If a, b, and c are multiples of 3 such that a > b > c > 0, which of the following values must be divisible by 3?

Indicate all such values.

A. a + b + c

B. a – b + c

C. \(\frac{abc}{9}\)

_________________

Indicate all such values.

A. a + b + c

B. a – b + c

C. \(\frac{abc}{9}\)

_________________

Sandy

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If you found this post useful, please let me know by pressing the Kudos Button

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Retired Moderator

Joined: **07 Jun 2014 **

Posts: **4810**

WE:**Business Development (Energy and Utilities)**

Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
17 Aug 2018, 16:00

4

Expert Reply

Explanation

Since a, b, and c are all multiples of 3, a = 3x, b = 3y, c = 3z, where x > y > z > 0 and all are integers.

Substitute these new expressions into the statements.

First statement: \(a + b + c = 3x + 3y + 3z = 3(x + y + z)\). Since (x + y + z) is an integer, this number must be divisible by 3.

Second statement: \(a - b + c = 3x - 3y + 3z = 3(x - y + z)\). Since (x + y + z) is an integer, this number must be divisible by 3.

Third statement:\(\frac{abc}{9}=\frac{3x \times 3y \times 3z}{9}=\frac{27xyz}{9}= 3xyz\). Since xyz is an integer, this number must be divisible by 3.

_________________

Sandy

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Since a, b, and c are all multiples of 3, a = 3x, b = 3y, c = 3z, where x > y > z > 0 and all are integers.

Substitute these new expressions into the statements.

First statement: \(a + b + c = 3x + 3y + 3z = 3(x + y + z)\). Since (x + y + z) is an integer, this number must be divisible by 3.

Second statement: \(a - b + c = 3x - 3y + 3z = 3(x - y + z)\). Since (x + y + z) is an integer, this number must be divisible by 3.

Third statement:\(\frac{abc}{9}=\frac{3x \times 3y \times 3z}{9}=\frac{27xyz}{9}= 3xyz\). Since xyz is an integer, this number must be divisible by 3.

_________________

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Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
06 Jul 2019, 13:36

sandy wrote:

If a, b, and c are multiples of 3 such that a > b > c > 0, which of the following values must be divisible by 3?

Indicate all such values.

A. a + b + c

B. a – b + c

C. \(\frac{abc}{9}\)

Indicate all such values.

A. a + b + c

B. a – b + c

C. \(\frac{abc}{9}\)

There is a little bit problem with this type of question, like when you solve the question. and get two or three answer choices, and when you just put your answer in one case (A), then try to put the second answer in (B/C) but before this, the GRE Timer just stop when you click answer A. Need some Web coding to dynamic this. Thanks

_________________

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Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
07 Jul 2019, 00:37

1

Expert Reply

I understand what do you mean. However, when you push the first answer you think is the timer stops.

This is fine. Ideally, when you hit the first answer and then the other(s) the workbook records them.

During the real exam is the same: you elaborate the question and then you do hit your answer(s).

Substantially, it does not change. The only difference is that here the timer stops and during the real exam continues.

Nonetheless, thank you for the feedback. If you think that it is an issue write on your own to the team about the timer.

https://gre.myprepclub.com/forum/ucp.php?i ... TimerIssue

Regards

_________________

This is fine. Ideally, when you hit the first answer and then the other(s) the workbook records them.

During the real exam is the same: you elaborate the question and then you do hit your answer(s).

Substantially, it does not change. The only difference is that here the timer stops and during the real exam continues.

Nonetheless, thank you for the feedback. If you think that it is an issue write on your own to the team about the timer.

https://gre.myprepclub.com/forum/ucp.php?i ... TimerIssue

Regards

_________________

Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
15 May 2020, 01:42

How is choice b not wrong?

If C=3, b=6, c=9

Then, 6+3-9=0

If C=3, b=6, c=9

Then, 6+3-9=0

Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
15 May 2020, 03:08

diegogre wrote:

How is choice b not wrong?

If C=3, b=6, c=9

Then, 6+3-9=0

If C=3, b=6, c=9

Then, 6+3-9=0

The condition states that, a>b>c

a=9, b=6, c=3

9-6=3 3+3=6

6/3 =3

Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
15 May 2020, 08:15

As embarrasing as it sounds. I didn't realize addition and subtraction are done left to right

Posted from my mobile device

Posted from my mobile device

Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
15 May 2020, 08:26

2

diegogre wrote:

As embarrasing as it sounds. I didn't realize addition and subtraction are done left to right

Posted from my mobile device

Posted from my mobile device

You're not the first person to believe that.

The acronym (PEMDAS or BEDMAS) seems to suggest that Addition must occur before Subtraction.

To avoid this problem, we should read the acronym as follows:

P

E

MD

AS

This format shows that multiplication and division are on the same hierarchical level. That is, neither operation is more important than the other. So, we perform these operations from left to right.

Likewise, the format shows that addition and subtraction are on the same level. So when given the option addition and subtraction, we perform those operations from left to right.

_________________

Re: If a, b, and c are multiples of 3 such that a > b > c > 0, w
[#permalink]
05 Mar 2024, 20:02

1

Addition and/or Subtraction of the multiples of a number will always yield another multiple of the number.

Therefore, A and B are correct

Now let us look at C.

The numerator abc will be a multiple of 3 and more specifically a multiple of 9 as each of these numbers contribute a 3 and 3 x 3 = 9. Thus abc is divisible by 9 and therefore by 3 as well.

Therefore C is also correct.

The correct answers are A, B, C.

_________________

Therefore, A and B are correct

Now let us look at C.

The numerator abc will be a multiple of 3 and more specifically a multiple of 9 as each of these numbers contribute a 3 and 3 x 3 = 9. Thus abc is divisible by 9 and therefore by 3 as well.

Therefore C is also correct.

The correct answers are A, B, C.

_________________

gmatclubot

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