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NOT a factor of M

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by Brent@GMATPrepNow » Tue Aug 30, 2016 10:08 am
Needgmat 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) 2100
E) 4900
OAA
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) 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.

Answer: 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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by [email protected] » Tue Aug 30, 2016 10:48 am
Hi Needgmat,

This question is essentially about prime-factorization. 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: A

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by Needgmat » Wed Aug 31, 2016 9:00 am
Hi Experts ,

Thank you so much for your explanation.

Just a quick question.

can you please explain why 700(selected randomly from choices) is not an answer, because if I expand 700, the I will get 2X2X5X5X7=700. In this there is no 3 so this can also be not a factor of M.

Please explain.

Many thanks in advance.

Kavin
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by Brent@GMATPrepNow » Wed Aug 31, 2016 10:51 am
Needgmat wrote:Hi Experts ,

Thank you so much for your explanation.

Just a quick question.

can you please explain why 700(selected randomly from choices) is not an answer, because if I expand 700, the I will get 2X2X5X5X7=700. In this there is no 3 so this can also be not a factor of M.

Please explain.

Many thanks in advance.

Kavin
700 = (2)(2)(5)(5)(7)

So, for 700 to be a factor of (2)(2)(3)(3)(5)(5)(7)(7), there must be two 2's, two 5's and one 7 "hiding" in the prime factorization (2)(2)(3)(3)(5)(5)(7)(7)

Here they are: (2)(2)(3)(3)(5)(5)(7)(7)

MORE EXAMPLES
Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
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by DavidG@VeritasPrep » Wed Aug 31, 2016 4:08 pm
For additional common multiple practice and a fun little discussion about a somewhat controversial question, see here: https://www.beatthegmat.com/question-on- ... 78608.html
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by Needgmat » Thu Sep 01, 2016 8:39 am

700 = (2)(2)(5)(5)(7)

So, for 700 to be a factor of (2)(2)(3)(3)(5)(5)(7)(7), there must be two 2's, two 5's and one 7 "hiding" in the prime factorization (2)(2)(3)(3)(5)(5)(7)(7)

Here they are: (2)(2)(3)(3)(5)(5)(7)(7)

MORE EXAMPLES
Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
Hi Brent ,

Thank you so much for your explanation. All clear now.

Thanks,

Kavin
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by Matt@VeritasPrep » Thu Sep 01, 2016 4:23 pm
Needgmat wrote:Hi Experts ,

Thank you so much for your explanation.

Just a quick question.

can you please explain why 700(selected randomly from choices) is not an answer, because if I expand 700, the I will get 2X2X5X5X7=700. In this there is no 3 so this can also be not a factor of M.

Please explain.

Many thanks in advance.

Kavin
Another way of seeing this: you know that the LCM is a multiple of 300, so it must divide by 100. You also know it's a multiple of 196, which means it's a multiple of 7, so it must divide by 7.

Since it divides by 7 and by 100, it must divide by 7*100, or 700, so B is out.
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