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

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If M is the least common multiple of 90,196, and 300, which [#permalink]

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27 Nov 2005, 06:43
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Feb 2014, 09:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.

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27 Nov 2005, 07:04
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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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27 Nov 2005, 07:20
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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25 Mar 2007, 07:04
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.

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25 Mar 2007, 10:18
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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 non-prime 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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25 Mar 2007, 21:38
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 [#permalink]

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01 Oct 2008, 01:28
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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

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

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01 Oct 2008, 12:19
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

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 [#permalink]

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01 Oct 2008, 14:47
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

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 [#permalink]

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24 Aug 2011, 01:40
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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 [#permalink]

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24 Aug 2011, 09:14
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 [#permalink]

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24 Aug 2011, 10:31
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 [#permalink]

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07 Feb 2014, 09:15
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 [#permalink]

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03 Jun 2014, 22:18
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

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

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17 Nov 2014, 05:28
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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 2^2
300 = 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

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

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30 Aug 2016, 10:12
Expert's post
Top Contributor
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)

A) 600
600 = (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.

[Reveal] Spoiler:
A

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 [#permalink]

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29 Nov 2016, 14:37
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 [#permalink]

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01 Dec 2017, 10:31
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Re: If M is the least common multiple of 90,196, and 300, which   [#permalink] 01 Dec 2017, 10:31
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