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

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

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)

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

the answer is A

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:

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:

Hi All,

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:

GMAT assassins aren't born, they're made,
Rich

Given that M is the least common multiple of 90,196, and 300 and we need to find which of the following is NOT a factor of M

Lets find the value of M first

90 = 2*3*3*5 = \(2^1 * 3^2 * 5^1\)
196 = 4*49 = \(2^2 * 7^2\)
300 = 2*2*3*5*5 = \(2^2 * 3^1 * 5^2\)

To find the LCM take all the terms and take their highest power in all of the above.

=> M = LCM(90, 196, 300) = \(2^2 * 3^2 * 5^2 * 7^2\) = 9 * 49 * 100
=> The Answer choices should divide the LCM

A. 600 Clearly, 600 is NOT a factor of the LCM as 600 = 6*100 and M = 9 * 49 * 100 and 9 * 49 is NOT DIVISIBLE by 6.
In exam we don't need to test further but I am solving to complete the solution.

B. 700 = 7 * 100. M = 441 * 100 and 9 * 49 = is divisible by 7 => POSSIBLE

C. 900 = 9 * 100. M = 441 * 100 and 9 * 49 is divisible by 9 => POSSIBLE

D. 2,100 = 21 * 100. M = 441 * 100 and 9 * 49 is divisible by 21 => POSSIBLE

E. 4,900 = 49 * 100. M = 441 * 100 and 9 * 49 is divisible by 49 => POSSIBLE

So, Answer will be A
Hope it helps!

To learn more about LCM and GCD watch the following videos

First we can break 90, 196, and 300 to primes to determine the LCM of those numbers:

90 = 9 x 10 = 2^1 x 3^2 x 5^1

196 = 4 x 49 = 2^2 x 7^2

300 = 30 x 10 = 2^2 x 3^1 x 5^2

Thus, the LCM is 2^2 x 3^2 x 5^2 x 7^2

Next, we can break down our answer choices into primes:

600 = 60 x 10 = 6 x 10 x 10 = 2^3 x 3^1 x 5^2

Since 600 has three 2s, it cannot be a factor of M.

Answer: A

I thought the question was about 2 number 90196 and 300. So there can be no space after the comma to separate different numbers? Or it is just a typo?

________________________
Typo. Edited. Thank you!

Asked: If M is the least common multiple of 90, 196, and 300, which of the following is NOT a factor of M?

90 = 2*3^2**5
196 = 2^2*7^2
300 = 2^2*3*5^2

M = LCM(90,196,300) = 2^2*3^2*5^2*7^2

A. 600 = 2^3*3*5^2 : NOT A FACTOR OF M
B. 700
C. 900
D. 2,100
E. 4,900

IMO A

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