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GRE Number Theory (Select One) http://www.800score.com/forum/viewtopic.php?f=12&t=3509 
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Author:  questioner [ Sat Mar 03, 2012 9:07 am ] 
Post subject:  GRE Number Theory (Select One) 
The least common multiple of positive integer m and 3digit integer n is 690. If n is not divisible by 3 and m is not divisible by 2, what is the value of n? A. 115 B. 230 C. 460 D. 575 E. 690 (B) We factorize 690 = 2 × 3 × 115 = 2 × 3 × 5 × 23. Dividing 690 starting by the smallest factor we get possible values for 3digit integer n: 690, 690 / 2 = 345, 690 / 3 = 230, 690 / 5 = 138, 690 / 6 = 115. 690 / 10 = 69 which is a 2digit number, therefore n must be one of the following: 690, 345, 230, 138, 115. n is not divisible by 3. So we eliminate 690, 345, 138. m is not divisible by 2, but the least common multiple of m and n is. So n must be divisible by 2. We eliminate 115. That leaves us only one option for n, 230. The answer is (B).  Hi, I understand why 690, 345, 138 are eliminated. But I do not understand why 115 was eliminated. Since this number is not divisible by 2 or 3. Where as 230 is divisible by 2 but not by 3. 
Author:  Gennadiy [ Sat Mar 03, 2012 9:22 am ] 
Post subject:  Re: t.4, qt.11: number theory, factorization 
Quote: But I do not understand why 115 was eliminated. Since this number is not divisible by 2 or 3. So we have 230 and 115 left.Now, forget about divisibility by 3. We've already used this fact to rule out the other 3digit divisors of 690. The question statement tells us "m is not divisible by 2". If n (the number we are looking for) was also not divisible by 2, then the LCM of m and n (690) would not be divisible by 2 as well. But 690 IS divisible by 2. So n must be divisible by 2. We rule 115 out because it is NOT divisible by 2. 
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