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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 ]

Re: A concept math, pls help [#permalink]
24 Aug 2011, 01:40

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)

Re: A concept math, pls help [#permalink]
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.

Re: A concept math, pls help [#permalink]
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. _________________

Consider me giving KUDOS, if you find my post helpful. If at first you don't succeed, you're running about average. ~Anonymous

Re: A concept math, pls help [#permalink]
24 Aug 2011, 17:16

Thank you so much, your expl. is very detailed and professional. I like yours.

I forgot this basic concept.

srivats212 wrote:

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)

Re: If M is the least common multiple of 90, 196, and 300, which [#permalink]
03 Jun 2014, 06:53

Hi Bunuel, Pls could you update the official answer here? Also I do it using systematically calculating LCM and prime factorizing each answer choice. But it takes long time. Is there possibly a shortcut? Thanks _________________

Please consider giving 'kudos' if you like my post and want to thank

Re: If M is the least common multiple of 90, 196, and 300, which [#permalink]
03 Jun 2014, 07:17

Expert's post

MensaNumber wrote:

Hi Bunuel, Pls could you update the official answer here? Also I do it using systematically calculating LCM and prime factorizing each answer choice. But it takes long time. Is there possibly a shortcut? Thanks

Merging similar topics.

No, finding the LCM and then checking the answer choices is pretty much it. _________________

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