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PhDmessi: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Mon Oct 03, 2011 1:33 pm; Thanked: 1 times

OG DS Arithmetic - please help with explanations

by PhDmessi » Mon Feb 27, 2012 11:16 am

Dear all,
I try to understand following question (Source: OG Quant, 1st Edition, DS 83) since yesterday, slept over it, still don't understand their explanation.

Q: If k and n are integers, is n divisible by 7?
(1) n-3=2k
(2) 2k-4 is divisible by 7

I solved so far, that neither (1) and (2) are sufficient by themselves.

Now OG says: Applying both (1) and (2), it is possible to answer the question. Through addition and subtraction, the equation n=2k+3 (yes, I got that) from (1) can be expressed as n=2k-4+7.
HOW? What did they add/subtract here? Why the -4? And they did not substitute the (2) so far.
Can please someone help me to understand what was done here or provide another way to solve the question?
Thanks a lot!

Never give up!

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Source: Beat The GMAT — Data Sufficiency | Mon Feb 27, 2012 11:16 am

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Mike@Magoosh: GMAT Instructor; Posts: 768; Joined: Wed Dec 28, 2011 4:18 pm; Location: Berkeley, CA; Thanked: 387 times; Followed by:140 members

by Mike@Magoosh » Mon Feb 27, 2012 11:59 am

Hi, there. I'm happy to help with this.

I'll show you two ways to think about this.

First of all, what the OG said. You see, 3 = 7 - 4, and because those are equal, we can substitute (7-4) for 3 in any context and it will still be equal.

When we know both statements,
(1) n-3=2k
(2) 2k-4 is divisible by 7

The first one is n = 2k + 3, and we would like to relate that to 2k - 4, about which we know some useful information. The OG uses the sly trick that 3 = (7-4), and says:

n = 2k + 3 = 2k + (7 - 4) = 2k - 4 + 7 = (2k - 4) + 7

Then, of course, 7 plus something divisible by 7 is still divisible by 7. This approach depends on seeing and applying that sly trick.

Here's another approach.

(1) n-3=2k ---> n = 2k + 3
(2) 2k-4 is divisible by 7

I'm going to write the second statement as 2k-4 = 7r, where r is some integer. Then,

2k = 7r + 4

n = 2k + 3 = [7r + 4] + 3 = 7r + 7

Therefore, n must be divisible by 7.

Does all of this make sense?

Here's another DS question on divisibility.

https://gmat.magoosh.com/questions/863

When you submit your answer to this question, it should be followed by a full video explanation of the solution.

Let me know if you have any further questions on this.

Mike

Magoosh GMAT Instructor
https://gmat.magoosh.com/

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PhDmessi: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Mon Oct 03, 2011 1:33 pm; Thanked: 1 times

by PhDmessi » Mon Feb 27, 2012 1:37 pm

Mike, thanks to your explanation I understood both ways! thx

So, this leaves me with following key takes aways/questions:
1. The main goal, should be to relate the two statements to each other. I tried that before, but I would never have used the method of manipulating a number (e.g: 3: -4 + 7) in this way (that's why I was so surprised I did not get a clue from the answer). I suppose, I should use it more often?

2. About your second way, that is exactly what I wanted to do when trying to solve the question; make the "(2) 2k-4 is divisible by 7" to an equation so that I can work with it such as 2k-4=7x but I did not know if this is mathematically correct. So is there any rule / suggestion for this type of questions?

Never give up!

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pm; Location: New York, NY; Thanked: 13060 times; Followed by:1906 members; GMAT Score:790

by GMATGuruNY » Mon Feb 27, 2012 2:00 pm

PhDmessi wrote:Dear all,
I try to understand following question (Source: OG Quant, 1st Edition, DS 83) since yesterday, slept over it, still don't understand their explanation.

Q: If k and n are integers, is n divisible by 7?
(1) n-3=2k
(2) 2k-4 is divisible by 7

Statement 1: n = 2k+3.
If k=1, then n=2(1)+3 = 5.
Is 5 divisible by 7?
NO.
If k=2, then n=2(2)+3 = 7.
Is 7 divisible by 7?
YES.
Since in the first case the answer is NO, but in the second case the answer is YES, INSUFFICIENT.

Statement 2: 2k-4 is divisible by 7.
No information about n.
INSUFFICIENT.

Statements 1 and 2 combined:
According to statement 2, 2k-4 = {7,14,21,28...}
Thus, 2k = {11,18,25,32...}
Since n=2k+3, n={14,21,28,35...}
In every case, n is a multiple of 7.
SUFFICIENT.

The correct answer is C.

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GMAT/MBA Expert

Mike@Magoosh: GMAT Instructor; Posts: 768; Joined: Wed Dec 28, 2011 4:18 pm; Location: Berkeley, CA; Thanked: 387 times; Followed by:140 members

by Mike@Magoosh » Mon Feb 27, 2012 3:29 pm

PhDmessi wrote:Mike, thanks to your explanation I understood both ways! thx

So, this leaves me with following key takes aways/questions:
1. The main goal, should be to relate the two statements to each other. I tried that before, but I would never have used the method of manipulating a number (e.g: 3: -4 + 7) in this way (that's why I was so surprised I did not get a clue from the answer). I suppose, I should use it more often?

Well, that trick, replacing a with (b - c), given that a = b - c, can be useful sometimes, but it is also sometimes hard to see when to use it. I would say, keep that tucked away in the back of your mind, especially in divisibility questions where things are being added. It's hard to say how often you will wind up using this.

PhDmessi wrote:2. About your second way, that is exactly what I wanted to do when trying to solve the question; make the "(2) 2k-4 is divisible by 7" to an equation so that I can work with it such as 2k-4=7x but I did not know if this is mathematically correct. So is there any rule / suggestion for this type of questions?

There is a useful general rule here. Suppose I divide D (called the "dividend") by P (called the "divisor"), and I get a quotient Q with a remainder R. This means:

D/P = Q + R/P, or in other words, D = P*Q + R

The dividend can be "rebuilt" from the divisor, the quotient, and the remainder. That is a very useful relationship to keep in mind on divisibility and factor questions.

I hope that's helpful. Let me know if you have any further questions.

Mike

Magoosh GMAT Instructor
https://gmat.magoosh.com/