Hold the laughter...question on Number Properties-Prime Box

[BT]

I know it's too early to be stumped, but here goes. In the Number properties guide - Page 17, question #5 asks the following:

If jis divisible by 12 and 10, is j divisible by 24?

I answered yes. My reasoning was as follows: The prime box for the number 12 contains 2,2, and 3. The prime box for the number 10 contains 2 and 5. My prime box for j therefore contains the collection of these prime factors - 2,2,2,3,5. If i multiply 2x2x2x3, i get 24.

The textbook, on the other hand, illustrates a prime box for j which contains 2,2,3, and 5. Since we can't get 24 by multiplying these factors, it says that the answer cannot be determined. I'm ok with the answer, but I'm trying to grasp why the prime box for j doesn't contain an additional 2. Am I overlooking something here? Your expertise would be appreciated.

Thanks[/b]

[shaji]

Certainly not laughing matter!!!. The LCM(Lowest Common Multiple) of 12 & 10=60. All odd multiples of 60 are not divisible by 24 while the even are. So NO/YES situation

[StaceyKoprince]

This is really tricky - a LOT of people struggle with these concepts!

When you are combining prime boxes, remember that you don't want to include overlapping information twice. So:
"12" box: 2, 2, 3
"10" box: 2, 5

combined: 2, 2, 3, (already have one 2 so don't include it again), 5

Think about it this way. You, me, and Shaji are standing outside of a store. I go inside, come back out, and tell you, "There are two oranges and an apple in there." (2=orange, 2=orange, 3=apple from our "12" prime box)

Then Shaji goes in, comes back out, and tells you, "There are an orange and a banana in there." (2=orange, 5=banana from our "10" prime box) (And yes that sentence is grammatically correct - there are - know your grammar rules too! :)

Are there definitely three oranges in the store? Or could Shaji and I have been talking about one of the same oranges? What's the minimum you know for sure? (Assuming you can trust us!). The minimum you know is that there are two oranges in there. There might be three. There might be five hundred. But all you know is that there are definitely at least two.

Same thing when doing the abstract "combine the prime boxes" - only include the minimum information that MUST be true and be aware that you have to strip out the potential overlap when combining separate prime boxes. Make sense?

[Guest]

Can you please explain how the explanation for Chapter 1, Question #5 from the Problem Set of Number Properties is different from the explanation provided for Question #1 (same chapter, same book).

1. If a is divided by 7 and by 18, an integer results. Is a/42 an integer?

5. If j is divisible by 12 and 10, is j divisible by 24?

I still want to say #5 is yes, based on the same reasoning used for Question #1.

Any assistance would be appreciated. Thank you.

[brian]

This question goes back to the core understanding of prime boxes.

What is the prime box for 12? 2,2,3
What is the prime box for 10? 2,5

12 & 10 Combined: 2,2,3,5

What is the prime box for 24? 2,2,2,3

So, clearly, not a match, right?

A great example would be the number #60 -- divisible by 10 & 12, but not by 24.

Hope that helps.

-Brian

[StaceyKoprince]

Remember, when combining prime boxes, you can only include the minimum that you definitely know - which means you have to strip out any overlap.

So when Brian combined the boxes for 12 and 10, note that he didn't end up with three 2's. Once I put the first two 2's in (from 12's prime box), I don't need to add another 2 from 10's box because I've got two 2's sitting in there already.

[blover]

Can you please explain how the explanation for Chapter 1, Question #5 from the Problem Set of Number Properties is different from the explanation provided under the sum and differnce are aslo divisible heading at page 14 (same chapter, same book).

5. If j is divisible by 12 and 10, is j divisible by 24?

adding them we got 22 and by sustracting them we got 2
prime box for 22 =2 x11
prime box for 2 =2 x1

[esledge]

The original question in this thread concerned the least common multiple (LCM) of 10 and 24.
10 = 2*5
24 = (2*2*2)*3
LCM = (2*2*2)*3*5
To get LCM quickly, you can list all the factors of both numbers: 2, 5, 2, 2, 2, 3
Then, cross off any "duplicates" (a single 2 appears on both lists, so cross one 2 off the list): 2--delete, 5, 2, 2, 2, 3
Take the product of what remains to get LCM.

The "sum and difference are also divisible" explanation on p. 14 has to do with the greatest common factor (GCF).
10 = 2*5
24 = (2*2*2)*3
GCF = 2
To get GCF quickly, take the product of the "duplicates" from one list to the other. Here, the only number that appears on both lists is 2.

Both explanations relate to the "duplicates" or shared factors of the two numbers, but it depends what you need to calculate:
GCF of x and y = the product of their shared factors
LCM of x and y = xy/the product of their shared factors = xy/GCF