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9 most common EMBA mistakes

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# M21 35

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Manager
Joined: 13 May 2010
Posts: 124
Followers: 0

Kudos [?]: 11 [0], given: 4

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14 Mar 2012, 10:04
If $$X$$ is a positive integer, what is the remainder of $$\frac{X}{3}$$ ?

1. The remainder of $$\frac{X}{4}$$ is 1
2. The remainder of $$\frac{X}{7}$$ is 6

Using number testing approach for testing these statements. When you combine these statements we need to come up with numbers that satisfy both the statements. I could come up quickly with 13....but could not figure out 41 was the other number.....what is a quick way of figuring out numbers that satisfy both constraints?
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 3523
Followers: 1190

Kudos [?]: 5314 [0], given: 58

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14 Mar 2012, 16:23
Hi, there. I'm happy to help with this.

The "meta-idea" behind this question is: numbers that are divisible by 4 are spaced 4 apart; numbers that when divided by 4 has a remainder of 1 (or 2 or 3), are also spaced four apart. Numbers that are divisible by 7 are spaced 7 apart; numbers that, when divided by 7 have any remainder you want to specify, are also spaced 7 apart. If you looking at numbers spaced four apart and numbers spaced 7 apart, you are not going to be lined up in any particular way with the numbers spaced 3 apart. Numbers spaced regularly by 4s or 7s will land all over the place on a sequence lined up by 3.

That's a very abstract way of saying it.

Concretely - - - -

Statement #1 --- X, when divided by 4, has a remainder of 1. Therefore, X could be 1, 5, 9, 13, 17, 21, 25, etc No conclusion about the remainder when dividing by 3. Statement #1, by itself, is insufficient.

Statement #2 --- X, when divided by 7, has a remainder of 6. Therefore, X could be 6, 13, 20, 27, 34, etc No conclusion about the remainder when dividing by 3. Statement #2, by itself, is insufficient.

Combined --- well, suppose we notice that 13 satisfies both conditions. How do we find other numbers that satisfy both conditions? Well, since we are dealing with patterns that repeat by 4 or 7 numbers, find the LCM of 4 and 7. The LCM of 4 and 7 is 28, so add 28

Numbers that satisfy both conditions included
13
13 + 28 = 41
41 + 28 = 69
69 + 28 = 97
etc.

13/3 --> r = 1
41/3 --> r = 2
69/3 --> r = 0
No conclusion about the remainder when dividing by 3. Both statements combined are insufficient. Answer =
[Reveal] Spoiler:
E

Does all this make sense?

Here's another practice question about divisibility
http://gmat.magoosh.com/questions/306
When you submit your answer to that question, the following page will have the full video solution.

Let me know if you have any questions on what I've said here.

Mike
_________________

Mike McGarry
Magoosh Test Prep

Manager
Joined: 28 Apr 2011
Posts: 195
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Kudos [?]: 8 [0], given: 6

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27 Mar 2012, 04:17
@mikemcgarry: Thanks a ton..........

Really good post.........

Re: M21 35   [#permalink] 27 Mar 2012, 04:17
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# M21 35

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