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Three is the largest number that can be divided evenly into

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Three is the largest number that can be divided evenly into [#permalink] New post 24 Sep 2012, 19:06
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Three is the largest number that can be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100

A. 30
B. 70
C. 210
D. 300
E. 700
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Sep 2012, 00:48, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Number systems doubt [#permalink] New post 24 Sep 2012, 20:56
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smartass666 wrote:
Three is the largest number that an be divided evenly into 27 and the positive integer x, while 10 is the largest number that can be divided evenly into both 100 and x. Which of the following is the largest possible number that could be divided into x and 2100
1 - 30
2- 70
3- 210
4 - 300
E - 700

I'm happy to help with this. :-)

This problem has to do with prime factorizations and Greatest Common Factor (GCF). You may find this blog article helpful:
http://magoosh.com/gmat/2012/gmat-math-factors/

27 = 3*3*3, and its GCF with x is 3, which implies that x has ONE factor of 3, but not TWO factors of 3.

100 and x have a GCF of 10, which implies that x has ONE factor of 10, but not TWO factors of 10.

Then we want to know what is the largest possible GCF of x and 2100.

Well 2100 = 3 * 7 * 10 * 10

We want x to include as many factors in common with 2100 as possible, to make the GCF with 2100 as big as possible.

We know x has one factor of 3, but not two factors --- that takes the 3.
We know x has one factor of 10, but not two factors --- we can take one of those 10's, but we have to leave the other
No other restrictions, so we can also grab that factor of 7 --- nothing saying that we can't, and it's there for the taking.

3*7*10 = 210

If we allow x to include as many factors as possible, within the constraints given, that is the most x could have in common with 2100.

Does all that make sense? Please let me know if you have any further questions.

Mike :-)
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Mike McGarry
Magoosh Test Prep

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Re: Number systems doubt [#permalink] New post 24 Sep 2012, 21:00
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ANSWER: C
Since 3 is the GCF of 27 (3.3.3) and x => x has only one "3 in its prime box.
Since 10 (2.5) is the GCF of 100 (2.2.5.5) and x => x has only one 2 and one 5 in its primebox.
So x = 2.3.5.n (n=something that is not 2, 3, 5)
2100 = 2.3.5.7
The GCF of x and 2100 is the product of the common factors of x and 2100 => the max possible is 2100.
Re: Number systems doubt   [#permalink] 24 Sep 2012, 21:00
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