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If M is the least common multiple of 90,196, and 300, which
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Updated on: 07 Feb 2014, 10:21
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If M is the least common multiple of 90,196, and 300, which of the following is NOT a factor of M? A. 600 B. 700 C. 900 D. 2,100 E. 4,900
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Originally posted by TOUGH GUY on 27 Nov 2005, 07:43.
Last edited by Bunuel on 07 Feb 2014, 10:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




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IF M is the least common multiple of 90,196 and 300, which of the following is NOT a factor of M?
A 600
B 700
C 900
D 2100
E 4900
90 = 2 * 3 * 3 * 5
196 = 2 * 2 * 7 * 7
300 = 2 * 2 * 3 * 5 * 5

LCM = 2 * 2 * 3 * 3 * 5 * 5 * 7 * 7
(TWO 2, TWO 3, TWO 5, TWO 7)
600 = 2 * 2 * 2 * 3 * 5 * 5
700 = 2 * 2 * 5 * 5 * 7
900 = 2 * 2 * 3 * 3 * 5 * 5
2100 = 2 * 2 * 3 * 5 * 5 * 7
4900 = 2 * 2 * 5 * 5 * 7 * 7
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If M is the least common multiple of 90, 196, and 300, which of the following is NOT a factor of M?
Rule of Thumb: whenever you see Least Common Multiple (LCM), think of factors and preferebly prime factorization. Every nonprime integer can be factored into only prime numbers. So lets start prime factoring:
[P.S. if you need help in prime factoring tell us]
For easy start, whenever you see an even integer, start with the prime factor 2
90: 3 x 30 > 3 x 3 x 10 > 3 x 3 x 2 x 5
196: 2 x 98 > 2 x 2 x 49 > 2 x 2 x 7 x 7
300: 2 x 150 > 2 x 2 x 75 > 2 x 2 x 3 x 25 > 2 x 2 x 3 x 5 x 5
LCM of the three number must contain all prime factors of each and every one of the three numbers : 90, 196, and 300
LCM : 2 x 2 x 3 x 3 x 5 x 5 x 7 x 7 = M
A factor of M must contain one or more, but limited to the available ones, of the prime factors of LCM [M]
A. 600 : 2 x 3 x 2 x 5 x 2 x 5 [ uses three 2's > Not a Factor ]
B. 700 : 2 x 5 x 2 x 5 x 7 [ a factor of M ]
C. 900 : 3 x 3 x 2 x 2 x 5 x 5 [ a factor of M ]
D. 2,100 : 7 x 3 x 2 x 5 x 2 x 5 [ you tell me ]
E. 4,900 : 7 x 7 x 2 x 5 x 2 x 5 [ yes indeedy ]
It is A



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Re: If M is the least common multiple of 90, 196, and 300, which
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01 Oct 2008, 02:28
90 = 3^2 x 2 x 5 196 = 7^2 x 2^2 300 = 3 x 5^2 x 2^2
so the least common = 7^2 x 2^2 x 3^2 x 5^2 = 7^2 x 3^2 x 100 = (7x3)^2 x 100
A. 6 = 3 x 2 <not B. 7 <ok C. 9 = 3^2 < ok D. 21 = 7 x 3 < ok E. 49 = 7^2 <ok
the answer is A



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Re: If M is the least common multiple of 90, 196, and 300, which
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17 Nov 2014, 06:28
Here is @lyla4's answer in a manner understandable to late bloomers like me LCM is the product of the greatest power of each prime that appears in any of the numbers. 90 = 3^2 x 2 x 5 196 = 7^2 x 2^2300 = 3 x 5^2 x 2^2 So the least common multiple is = 7^2 x 2^2 x 3^2 x 5^2 = (7x3)^2 x 100 Divide each answer choice and the LCM by 100 Therefore LCM = (7x3)^2 Now we are looking for the choice in which, LCM/choice is not an integer. A. 6 = 3 x 2 <not B. 7 <ok C. 9 = 3^2 < ok D. 21 = 7 x 3 < ok E. 49 = 7^2 <ok the answer is A
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Re: If M is the least common multiple of 90,196, and 300, which
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29 Jan 2018, 15:45
Hi All, This question is essentially about primefactorization. Here's a simple example of that concept: What is the least common multiple of 10 and 15. Now you probably already know that the LCM is 30, but here's WHY it's 30... 10 = (2)(5) 15 = (3)(5) When looking for the LCM, we need to multiply all of the prime factors of the numbers involved. However, each instance of a prime that shows up in both numbers should be counted just once (here, there's one 5 in both numbers, so we count that as just ONE 5 and not two 5s). This gives us... (2)(3)(5) = 30 We can then use those primes to figure out all of the divisors of the LCM: 1 2 3 5 (2)(3) = 6 (2)(5) = 10 (3)(5) = 15 (2)(3)(5) = 30 The exact same concept applies to this question  it's just that there's a lot more math work involved: 90 = (2)(3)(3)(5) 196 = (2)(2)(7)(7) 300 = (2)(2)(3)(5)(5) The LCM of these three numbers will include two 2s, two 3s, two 5s and two 7s: (2)(2)(3)(3)(5)(5)(7)(7) At this point, you should NOT multiply those numbers together  we're just going keep them as a reference point so that we can find the one answer that is NOT a possible factor from that list: Let's start with the easiest option first: Answer A: 600 = (2)(2)(2)(3)(5)(5) Notice how this number hast THREE 2s. This number is NOT possible given the list of primes that we have to work with, so it cannot be a factor of M. Final Answer: GMAT assassins aren't born, they're made, Rich
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Yes, got the same ans.
M = 2*2*3*3*5*5*7*7
600 = 2*2*2*5*5*3  one 2 is left out here. So that is the answer.



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first calculate the LCM of the given numbers
90 = 2*3*3*5
196=2*2*7*7
300=2*2*3*5*5
LCM = 2*2*3*3*5*5*7*7 this is the number M.
now check each number whether a factor of M.
Answer is 600.



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Thank you vshaunak and Mishari for your clear explanation!!
So this is how to approach this kind of problem.
I should learn more of these approaches so that I don't have to plug in numbers for number property probs!



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Re: If M is the least common multiple of 90, 196, and 300, which
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01 Oct 2008, 13:19
lylya4 wrote: 90 = 3^2 x 2 x 5 196 = 7^2 x 2^2 300 = 3 x 5^2 x 2^2
so the least common = 7^2 x 2^2 x 3^2 x 5^2 = 7^2 x 3^2 x 100 = (7x3)^2 x 100
A. 6 = 3 x 2 <not B. 7 <ok C. 9 = 3^2 < ok D. 21 = 7 x 3 < ok E. 49 = 7^2 <ok
the answer is A I think, you missed one 2 in the least common multiple. It will be 7^2 * 5^2 * 3^2 * 2^3. To me, all the answer choices are factors of this least common multiple. Is there any typo in the question?



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Re: If M is the least common multiple of 90, 196, and 300, which
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01 Oct 2008, 15:47
scthakur wrote: lylya4 wrote: 90 = 3^2 x 2 x 5 196 = 7^2 x 2^2 300 = 3 x 5^2 x 2^2
so the least common = 7^2 x 2^2 x 3^2 x 5^2 = 7^2 x 3^2 x 100 = (7x3)^2 x 100
A. 6 = 3 x 2 <not B. 7 <ok C. 9 = 3^2 < ok D. 21 = 7 x 3 < ok E. 49 = 7^2 <ok
the answer is A I think, you missed one 2 in the least common multiple. It will be 7^2 * 5^2 * 3^2 * 2^3. To me, all the answer choices are factors of this least common multiple. Is there any typo in the question? A. LCM contains 2 2's whereas 600 = 25 * 3 * 8. thus cannot be a factor.



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Re: A concept math, pls help
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24 Aug 2011, 02:40
The key to this problem is breaking down the numbers into its prime factors. After that its a piece of cake!
90 > 3*3*5*2 196 > 2*2*7*7 300 > 3*2*2*5*5
M(LCM) = multiply all the factors (pick the highest power of the common factor)
M(LCM) = \(2^2*3^2*5^2*7^2\)
The question asks which is NOT A FACTOR of M: Only A(600) which has an additional factor 2 is not a factor of M 600 > \(2^3*3*5^2\) => it has an additional factor of 2 which is not present in M(only 2 factors of '2' is present here)



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Re: A concept math, pls help
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24 Aug 2011, 10:14
Just because I've into solving multiple problems today.
Here's how I think of it: you essentially need to pull out the primes one by one from 900, 196 and 300 respectively. The easiest way for me to do this is pull out tens first (which consists of 2*5, both primes) so 900  2 * 5 * 2 * 5 , leaving a 9 which is 3 *3 so you have 2, 2, 3, 3, 5, 5 300  2 *5 * 2 * 5, leaving 3 so 2, 2, 3, 5, 5 196 is trickier. Up until today I didn't know it was the square of 14 but once you know that: 196 = 2 * 7 * 2 * 7 so 2, 2, 7, 7 then make a list like srivats212 said but including the the number as much as it appears in any given number so 2, 2, 3, 3, 5, 5, 7, 7
Now find the number that requires any prime number to pop up more than it does in the bolded list.
600 = 2 * 5 * 2 * 5 * 2 * 3, so it needs 3 2's and you only have 2 twos in the bolded list, so isn't a factor of M.
I think if you want to do these things fast it's best to memorize all the squares up to 20. Which I haven't bothered up until now either, just did this in excel. 11 121 12 144 13 169 14 196 15 225 16 256 17 289 18 324 19 361 20 400 These numbers alone or multiples of these numbers seem to pop up a lot I suppose in other types of problems too. But for multiple/factor problems all you need to do is figure out the prime factors of the original number. For 324, the square root is 18. Then prime factors of 324 are 2 * 9 * 2 * 9. But memorization of the squares is essential for pattern recognition (and hence completion of problems in 2 minutes or less) I'm starting to realize.



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Re: A concept math, pls help
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24 Aug 2011, 11:31
A should the answer. We can get if by combined factorization. 600 is the only # that we can't get from combined factorization to calculate LCM.
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Re: IF M is the least common multiple of 90,196 and 300, which
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07 Feb 2014, 10:15
90 = 2 * 3^2 * 5 196 = 2^2 * 7^2 300 = 2^2 * 3 * 5^2 LCM contains the HIGHEST power of EVERY number present in prime factorization. M = LCM = 2^2 * 3^2 *5^2 * 7^2.
Look @ answer options. A: 600 = 2^3 * 3 * 5^2 >Cannot be a factor because of 2^3.



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Re: If M is the least common multiple of 90,196, and 300, which
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03 Jun 2014, 23:18
90 = 2 * 3 * 3 * 5 196 = 2 * 2 * 7 * 7 300 = 2 * 2 * 3 * 5 * 5 Only 2 is common in all the 3 factorization So, LCM =\(\frac{90 * 196 * 300}{8} = 45 * 49 * 300\)(No need of further calculation) Testing Option I > 600. It is NOT the factor Answer = A = 600
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Re: If M is the least common multiple of 90,196, and 300, which
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30 Aug 2016, 11:12
TOUGH GUY wrote: If M is the least common multiple of 90,196, and 300, which of the following is NOT a factor of M?
A. 600 B. 700 C. 900 D. 2,100 E. 4,900 The least common multiple of 90, 196, and 300 = (2)(2)(3)(3)(5)(5)(7)(7) So, M = (2)(2)(3)(3)(5)(5)(7)(7) Now check the answer choices... A) 600600 = (2)(2)(2)(3)(5)(5) For 600 to be a factor of M, there must be three 2's, one 3 and two 5's "hiding" in the prime factorization of M. Since, M only has two 2's in its prime factorization, 600 is NOT a factor of M. Answer: For more on the relationship between factor and prime factorization, watch this video: https://www.gmatprepnow.com/module/gmat ... /video/825 ASIDE: I thought it might be useful to show one way to find the LCM of large numbers:
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Re: If M is the least common multiple of 90,196, and 300, which
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29 Nov 2016, 15:37
90 = (2^3)(3^2)(5)
196 = (2^2)(7^2)
300 = (2^2)(3)(5^2)
LCM > (2^2)(3^2)(5^2)(7^2)
A. is the correct answer > 600 cannot be made from the prime numbers found in the LCM shown above.



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Re: If M is the least common multiple of 90,196, and 300, which
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