MGMAT, FACTORS

well the simplest way is to realize the set..

first set d is all multiples of 3 between 20 and 100...
let them whatever they r be...
they are some numbers in between 20 and 100...

second
the set # will contain factors of 400...
now id u see the prince factorization of 400 = 4^2 * 5^2

it will never have the factor of 3 in its factors...

so by no chance can any multiple of 3 between 20 and 100 be a factor of 400...

so there will be no common numbers...

Hence pick A,

Hope this helps..

Unfortunately, it isn't clear to me yet :'(

Could someone please explain further for the benefit of everyone ?

neerajkumar1_1 wrote:well the simplest way is to realize the set..

first set d is all multiples of 3 between 20 and 100...
let them whatever they r be...
they are some numbers in between 20 and 100...

second
the set # will contain factors of 400...
now id u see the prince factorization of 400 = 4^2 * 5^2

it will never have the factor of 3 in its factors...

so by no chance can any multiple of 3 between 20 and 100 be a factor of 400...

so there will be no common numbers...

Hence pick A,

Hope this helps..

Hey pesfunk:

Good question - let me see if I can help!

One of my favorite things about problems like this is that they so often clue you in to the fact that you need to think about factors by having the word "factor" in the problem itself. We're asked to find numbers in common between:

The multiples of 3 between 20 and 100

and

The factors of 400


What's important to note here is that any positive integer that is not prime can be expressed as a product of its prime factors. (And I apologize for the awkwardness of that statement in the name of precision)

So, for 400, we can break it down into:

4 * 100

2*2 * 2*50

2*2 * 2 * 2*25
2*2* 2* 2* 5 *5

So we know that the factors of 400 are, when stripped down to their essentials, products of 2s and 5s.


With the multiples of 3 between 20 and 100, they'll start at:

21, 24, 27, 30, etc.

so we can break those down into:

3*7; 3*8; 3*9; 3*10

Now, what's important to note here is that EACH MULTIPLE OF 3 HAS A PRIME FACTOR OF 3.

But for the factors of 400, none of them have a factor of 3.

So there isn't any room for overlap - they won't share any numbers in common because set D requires each number to have a prime factor of 3 in it, and none of the numbers in set E have that.




Now, more intuitively, you can ask yourself: How many multiples of 3 are factors of 400? And because 400 isn't divisible b y 3, the answer is "none".

Or, if you want to just start listing out, you know that to match with a number in D you'd have to be less than 100, so we can just list the factors of 400 that are less than 100:

1, 2, 4, 5, 8, 10, 20, 40, 50, 80

None of those are multiples of 3, so the answer again is "none".