Foundations of Math

[kouranjelika]

Hey guys,

First time posting a topic on here, so I hope it's in the right section and I'm following all the instructions on postings.

So although scoring above 15 on the diagnostics, I decided to review all the Foundations of Math before starting my course next week.

I was doing the drills and got one question wrong. I watched the accompanying workshop and understand the basis behind the explanation, however still do NOT fully follow why the answer is, what it is.

Foundations of Math I
Games Drill
Question 2.

If x is divisible by 6 and 9, then x must be a multiple of
I. 3 II. 18 III. 27
(A) I only
(B) I and II only
(C) II and III only
(D) I and III only
(E) I, II and III

So the answer is (B), but I don't completely follow why it could not be III as well, hence answer (E).

I understand that the primes of 6 are 2,3; 9 are 3,3.
Hence the minimal X would be 3X3X2; hence 18 and a multiple of I. 3*6 II. 18*1.
BUT, if x would be a multiple of any of the numbers presented, and if x is for example 18*3 (which is still divisible by 6&9), 54 would also be a multiple of 27.
Can you elaborate on that please?

Thank you,
Anjelika

[georgepa]

The word "must" in the question is the important limiting factor. Sure x=54 is divisible by 27, 18, and 3. However, by your own logic, if x=18, x is certainly is not divisible by 27. For the choices given - only 3 and 18 "must" be divisors of any x - given that 6 and 9 are also divisors of x.


Prime Boxes

Code: Select All Code

X has divisors 6, and 9 (and possibly others  ->? )
  6         
[2,3,?]  => At least one 2 and at least one 3

  9
[3,3,?] => At least two 3s

So combining we know that X has divisors [2,3,3,?]  => At least one 2 and at least two 3s


Since we don't know what ? is - the only facts we do know ("must" know) are that x has the following factors

1. 2 -> Not in the answer choice


2. 3 -> Choice I


3. 2 x 3 = 6 -> given


4. 3 x 3 = 9 -> given


5. 2 x 3 x 3 = 18 -> Choice II

So (B) I and II only

[kouranjelika]

Hey George,

But we DO NOT know that x is 18. x is simply 18x, meaning it is SOME multiple of 18, not necessarily 1. In which case, it may very well be 27 as I said, can you back up the reasoning behind only using 18*1 and not 18*3 or some other 18 multiple?
Why only 1 is used in the solution of this fairly simple question is what I am wondering?

Thank you,
Anjelika

[georgepa]

Again - its the "must" in the question that is important.

If x = 18n

What must be true? What can you say about x without knowing anything about n that has to be true?

Another way of reasoning this is the following:

Since you don't know anything about n in x=18n, the only thing you can say with any certainty about x comes from the 18 contained in x.

Testing cases will make it apparent (I've italicized 6 and 9 which are already given and marked them at the end of the list)

n = 1 => x = 18 => factors are 1, 2, 3, 18, 6, 9
n = 2 => x = 36 => factors are 1, 2, 3, 4, 12, 18, 36, 6, 9
n = 3 => x = 54 => factors are 1, 2, 3, 18, 27, 54, 6, 9
n = 4 => x = 72 => factors are 1, 2, 3, 4, 8, 12, 18, 24, 36, 72, 6, 9
n = 5 => x = 90 => factors are 1, 2, 3, 5, 10, 15, 18, 30, 45, 90, 6, 9

You can see that only 1,2,3,6,9, and 18 have to be factors in any value of x - including the case when n=1 - which is a valid x. That is for any x - you will find that any number in the set {1,2,3, 6,9,18} MUST DIVIDE x

Another way of looking at it: The factor pairs of 18 are
1 18
2 9
3 6

These MUST divide x no matter what x is, since x = 18n

[kouranjelika]

I believe I follow now and hope I can apply this understanding to a similar problem as it comes up.

Thank you so much for your time George.

Anjelika

[RonPurewal]

To the original poster"”

* This is, in fact, the wrong folder for this question. The foundations book is a MGMAT source, so this belongs in the "MGMAT non-CAT math" folder.

* The post is also titled incorrectly. The title of the post should contain the first few words of the problem statement.

Please read the forum rules (first post in every folder) before posting. Thanks.

[RonPurewal]

Also, there's a much, much, MUCH easier way to solve this problem than to think about the theory of divisibility and primes. You can just make a list of numbers that satisfy the given criterion, and then go from there.

I.e., just think of numbers that are divisible by 6 and by 9.
The first such number is 18. Then, 36. Then, 54. Etc.
These are clearly all multiples of 3 (which we would already know just from the 6 or the 9 alone). A little investigation will reveal that they're all multiples of 18.

As soon as you have the very first number in the whole list (18)"”which is not a multiple of 27"”you can eliminate 27 and never look at it again.

Since this problem is in an incorrect folder, I'm going to lock the thread. If you have further questions, please post a new thread in the correct folder.
Thanks.