Foundations of Math Divisibility Question (HW Banks)

[Milanproda1]

Hello,

I am having trouble understanding this question, and while the book does a great job explaining things, I seem to be unable to grasp the concept:

If x is divisible by 6 and 9, then x must be a multiple of

I. 3
II. 18
III. 27

I only
I and II only
II and III only
I and III only
I, II, and III
The prime factors of 6 are 2 and 3. So x must be divisible by at least one 2 and at least one 3.

The prime factors of 9 are 3 and 3. So x must be divisible by at least two 3's.

Putting these two pieces of evidence together, we conclude that the prime factors of x include at least one 2 and at least two 3's. Thus, x must be divisible by 2×3×3, which equals 18. Moreover, x also has to be divisible by all the factors of 18, including 3. However, there is no need for x to be divisible by 27 (=3×3×3), since we don't know whether x contains a third 3.

The correct answer is B.

If X is divisible by both, 6 and 9, would X not have one 2 and three 3's (6=2x3 & 9=3x3)?

I chose I II and III, and at first it did not make sense, but when I did my amateur version of the math, 27 seemed to fit in as well.

Thanks ahead of time,

Milan

[jnelson0612]

Milan,
You nicely worked out that x must be a multiple of 18. x definitely contains one 2 and two 3s, which results in a product of 18. x may have other factors, but we know it has those. Yes to statement II.

Also, if x is a multiple of 18, it must also be a multiple of each of 18's factors. Thus, since 3 is a factor of 18, x must also be a multiple of 3. Yes to statement I.

However, does x have to be a multiple of 27? No. Based on the information we have, x could be 18, which is not a multiple of 27. If we knew that we have a third 3 then we would know x is a multiple of 27, but since we don't have that knowledge we cannot definitively conclude that statement III is true.

I hope this answers your questions.

[Milanproda1]

Jaime,

Thank you very much for your reply!

I am having a great deal of confusion in regards to the number of three's in this question. Please bear with me as I try to comprehend the info.

My understanding is that N would contain one 2 and three 3's: 2X3=6 and 3X3=9.

27 also has three 3's (3^3). In order for for X to be a multiple of 27, it would have to have all of the same prime factors as 27, which (under my logic) it does.

You made a great point about X possibly being 18, which proves that I am incorrect. However I am not sure where my logic was wrong.

Regards,

Milan

[jnelson0612]

My understanding is that N would contain one 2 and three 3's: 2X3=6 and 3X3=9.

You made a great point about X possibly being 18, which proves that I am incorrect. However I am not sure where my logic was wrong.

Hi Milan,
I have bolded the part in which your logic is leading you astray.

We know that x is divisible by 6 and 9. You are correct about the factors:
6 contains one 2 and one 3
9 contains two 3s

However, you are counting each 3 as separate and that is leading you to conclude that there are three 3s. The reality is that one of the 3s in 9 may be the exact same 3 that is in 6. We *really* only know that we have at least two 3s.

One way to illustrate this is to draw two circles that have a place of overlap in the middle. We know that 6 and 9 each have one 3 in common. Place that in the overlapping part. Then place a 2 in one of the circles (let this circle represent 6) and a 3 in the other circle (let this circle represent 9). By multiplying across (2 * 3 * 3) we can obtain our least common multiple, 18.

Let me put it another way. Let's say you and I are sitting in a room together. A man enters the room and says that he just saw three elephants walk by on the street. Then a woman enters the room and says that she just saw two elephants walk by. Can we conclude that there are at least five distinct elephants? We can't. We can only conclude that there are at least three, since the two the woman saw may be part of the three the man saw. The same principle applies to the counting of the 3s in this problem.

I hope this helps. Please let me know if I can provide further clarification.