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# Three is the largest number that can be divided evenly into

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Joined: 11 Feb 2012
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Three is the largest number that can be divided evenly into [#permalink]  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.
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 2787
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Kudos [?]: 3960 [7] , given: 44

Re: Number systems doubt [#permalink]  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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Schools: HSG '15 (A)
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Kudos [?]: 35 [4] , given: 34

Re: Number systems doubt [#permalink]  24 Sep 2012, 21:00
4
KUDOS
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.
Manager
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Kudos [?]: 23 [0], given: 78

Three is the largest number that can be divided evenly into 27 [#permalink]  12 Oct 2015, 16:43
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
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Manager
Joined: 04 Feb 2014
Posts: 53
WE: Project Management (Manufacturing)
Followers: 0

Kudos [?]: 23 [0], given: 78

Three is the largest number that can be divided evenly into 27 [#permalink]  12 Oct 2015, 16:50
Three is the largest number that can be divided evenly into 27 and the positive integer x
So x contains 3

10 is the largest number that can be divided evenly into both 100 and x
So x contains 2 and 5

So x = 2.3.5.n

largest possible number that could be divided into x and 2100
2100= 3*2*2*5*5*7
x = 2.3.5.n

So largest no should be 300=3*5*5*2*2 as it contains 2,3 and 5

What am I doing wrong???
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Kudos [?]: 1575 [0], given: 152

Re: Three is the largest number that can be divided evenly into 27 [#permalink]  12 Oct 2015, 16:56
Expert's post
Hi anurag16,

This question can be dealt with in a couple of different ways, but it ultimately comes down to Prime Factorization.

We're told that the largest number that divides into 27 and X is 3.

Since 27 = (3)(3)(3), that means that X's prime factorization can contain JUST ONE 3 (although it can contain other prime factors). If it contained more than one 3, then "3" would NOT be the largest number that would divide into 27 and X.

So X could be 3, 6, 12, 15, 30, etc.

Next, we're told that the largest number that divides into 100 and X is 10.

Since 100 = (2)(2)(5)(5), that means that X's prime factorization can contain JUST ONE 2 AND JUST ONE 5 (although it can contain other prime factors).

So we know that X's prime factorization consists of ONE 2, ONE 3 and ONE 5 and possibly some other primes.

We're asked for the LARGEST number that could divide into X and 2100.

2100 = (3)(7)(2)(2)(5)(5). Using what we know about X, the LARGEST number that could divide X AND 2100 would be (2)(3)(5)(7) = 210.

[Reveal] Spoiler:
C

GMAT assassins aren't born, they're made,
Rich
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# Rich Cohen

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# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6216 Location: Pune, India Followers: 1674 Kudos [?]: 9583 [1] , given: 196 Re: Three is the largest number that can be divided evenly into [#permalink] 12 Oct 2015, 22:12 1 This post received KUDOS Expert's post smartass666 wrote: 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 This question is based on very simple concepts but the wording is a bit sneaky. We will come to that in a minute. First, let's quickly analyse what we are given: 3 is the largest number that can be divided evenly into 27 ( = 3^3) and the positive integer x --> 27 has three 3s but only one 3 is common to 27 and x. It means x has exactly one 3. 10 is the largest number that can be divided evenly into both 100 (= 2^2 * 5^2) and x --> Only one 2 and one 5 is common between 100 and x so x has exactly one 2 and exactly one 5. So x has exactly one 3, one 2 and one 5. Which of the following is the largest possible number that could be divided into x and 2100? We need to find the largest possible number that could be divided in x and 2100. Note that x cannot have more than one 3, 2 and 5. 2100 = 2^2 * 3 * 5^2 * 7 It has two 2s and two 5s. But x cannot have two 2s and two 5s - these are two of the limiting conditions we have on x. What about 7? The questions doesn't tell us whether x has 7. But could it have 7? Sure. We have no limiting condition on having other factors. So the maximum common factors that 2100 and x could have are 2 * 3 * 5 * 7 = 210 So 210 is the largest possible number that could be divided into x and 2100. Answer (C) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Three is the largest number that can be divided evenly into   [#permalink] 12 Oct 2015, 22:12
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