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Three is the largest number that can be divided evenly into [#permalink]
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24 Sep 2012, 20:06
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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
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Last edited by Bunuel on 25 Sep 2012, 01:48, edited 1 time in total.
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Re: Number systems doubt [#permalink]
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24 Sep 2012, 21:56
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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/gmatmathfactors/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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Re: Number systems doubt [#permalink]
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24 Sep 2012, 22:00
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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.



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Three is the largest number that can be divided evenly into 27 [#permalink]
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12 Oct 2015, 17:43
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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Three is the largest number that can be divided evenly into 27 [#permalink]
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12 Oct 2015, 17:50
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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Re: Three is the largest number that can be divided evenly into 27 [#permalink]
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12 Oct 2015, 17:56
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. Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: Three is the largest number that can be divided evenly into [#permalink]
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12 Oct 2015, 23:12
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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)
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Three is the largest number that can be divided evenly into [#permalink]
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22 Mar 2016, 00:22
This question is tricky because of the convoluted language. So x has one 3, one 5 and one 2 as prime factors, and also could have some other primes as no contraints have been set: x = 3*5*2*n. That is 30, 30*7, 30*11, 30*17... 2100 = 7*3*5*2*5*2. These two numbers obviously share such primes as 3*5*2. Since x cannot have more 3s, 2s, and 5s as prime factors we can borrow one 7 from 2100 to construct our max possible common factor of the 2100 and x = 30*7=210.
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