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Basic Prime Box Question

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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by Maciek » Sun Aug 29, 2010 6:28 am
Hi Noah!

you should watch this lesson:
https://www.beatthegmat.com/mba/2010/06/ ... rime-boxes

let us create DS question:
DS. Is j divisible by 24?
(1) j is divisible by 12
(2) j is divisible by 10
(1) we have divisibility statement: j is divisible by 12
From the first statement we know that j must contain 2, 2 and 3.
if j were divisible by 24, it would contain 2, 2, 2 and 3.
so this statement ALONE is INSUFFICIENT

(2) we have divisibility statement: j is divisible by 10
From the second statement we know that j must contain 2 and 5.
if j were divisible by 24, it would contain 2, 2, 2 and 3.
so this statement ALONE is INSUFFICIENT

(1) & (2) we have 2 divisibility statements: j is divisible by 12 AND j is divisible by 10
prime box of j includes digits 2, 2, 2, 3 and 5
Image
so answer is yes
so both statements TOGETHER are SUFFICIENT

IMO C

Hope it helps!
Best,
Maciek
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by neerajkumar1_1 » Mon Aug 30, 2010 6:54 am
Maciek wrote:Hi Noah!

you should watch this lesson:
https://www.beatthegmat.com/mba/2010/06/ ... rime-boxes

let us create DS question:
DS. Is j divisible by 24?
(1) j is divisible by 12
(2) j is divisible by 10
(1) we have divisibility statement: j is divisible by 12
From the first statement we know that j must contain 2, 2 and
if j were divisible by 24, it would contain 2, 2, 2 and 3.
so this statement ALONE is INSUFFICIENT

(2) we have divisibility statement: j is divisible by 10
From the second statement we know that j must contain 2 and 5.
if j were divisible by 24, it would contain 2, 2, 2 and 3.
so this statement ALONE is INSUFFICIENT

(1) & (2) we have 2 divisibility statements: j is divisible by 12 AND j is divisible by 10
prime box of j includes digits 2, 2, 2, 3 and 5
Image
so answer is yes
so both statements TOGETHER are SUFFICIENT

IMO C

Hope it helps!
Best,
Maciek


I suppose u got the answer wrong...

statement 1: IF j is divisible by 12 then it will have 2 2 3
insufficient, as if j were to be divisible by 24 then it shud have 2 2 2 3

statement 2: IF j is divisible by 10 then it will have 2 5
insufficient, as if j were to be divisible by 24 then it shud have 2 2 2 3

together

u take the common factors and not just multiply the terms

so if j is divisible by 10 and 12... then j will have 2 2 3 5

since j still does not have 2 2 2 ( 2^3) it will not be always be divisible by 24....


Hence answer is E
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by Maciek » Mon Aug 30, 2010 7:24 am
Hi!
if j is divisible by 10 and 12... then j will have 2 2 3 5
2*2*3*5 = 60
60 is divisible by 10 and 12 but not by 24
You're right.
this is correct prime box:
Image
Best,
Maciek
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by oex85 » Wed Jul 06, 2011 8:48 am
Hi. Been lurking around for a couple of weeks here. Great site! I'm hijacking this thread instead of creating a new thread...

I'm interested in the same question as above, which is in MGMAT:s number properties book, but in relationship to a similar (I think) question in the same problem set.

For the above question I got two individual prime boxes [2,2,3] and [2,5]. Maybe there's something I'm not getting, because I can't see why j is not divisable by 24 if you can create 24 out of the numbers in the two boxes. "The 2 in the prime factorization of 10 may be REDUNDANT", it says in the book. What does this mean?

And what's the difference between that question and:

"If 24 is a factor of h and 28 is a factor of k, must 21 be a factor of hk?"

Is it because it is now "factors" instead of "divisable"? Because I did exactly the same with prime boxes and combined 3 from the first with 7 from the second...

Appreciate any help understanding the nuances of this whole divisibility & primes chapter. Cheers.
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by Tani » Fri Jul 08, 2011 1:28 pm
When combining prime boxes you include each factor the maximum number of times it appears in any individual box. Therefore, for 10 and 12 you only need two 2s one 3 and one 5.

Check that 60 is divisible by 10 and 12, but not 24.
Tani Wolff
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by oex85 » Sat Jul 09, 2011 8:58 am
Tani Wolff - Kaplan wrote:When combining prime boxes you include each factor the maximum number of times it appears in any individual box. Therefore, for 10 and 12 you only need two 2s one 3 and one 5.

Check that 60 is divisible by 10 and 12, but not 24.
Thank you for your answer, Tani. My post was a mess though and I probably should have posted it separately.

The question I'm wondering about is really this one:

If 24 is a factor of h and 28 is a factor of k, must 21 be a factor of hk?

Prime boxes: 24: [2,2,2,3] and 28: [2,2,7].

Answer: By the Factor Foundation Rule, all the factors of both h and k must be factors of the product, hk. Therefore, the factors of hk include, 2,2,2,2,2,3 and 7, as shown in the combined prime box. 21=3x7. Both 3 and 7 are in the prime box. Therefore 21 is a factor of hk.


What's the difference between these two questions? It seems to me that if you can just take 3x7 to get 21, then why can't you use 2x2x3x2 to get 24?

Hope I'm clearer this time around.

Appreciate it!
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by Tani » Sat Jul 09, 2011 11:15 am
The problem is not using 2 and 3, the problem is using too many 2s. Your example of 28 and 24 is asking a different question. It wants to know what must be a factor of the common multiple, not what is the least common multiple.

Any multiple of 10 will have AT LEAST one 2. Any multiple of 12 will have AT LEAST two 2s. A common multiple only needs the maximum number of 2s in EITHER of the factors. Neither one requires three 2s so the multiple doesn't REQUIRE three 2s. There are certainly many common multiples that will include three 2s, but it is not a requirement. When you are looking for a LEAST common multiple you have to use the LEAST possible number of each prime factor.
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by oex85 » Sat Jul 09, 2011 5:28 pm
Tani Wolff - Kaplan wrote:The problem is not using 2 and 3, the problem is using too many 2s. Your example of 28 and 24 is asking a different question. It wants to know what must be a factor of the common multiple, not what is the least common multiple.

Any multiple of 10 will have AT LEAST one 2. Any multiple of 12 will have AT LEAST two 2s. A common multiple only needs the maximum number of 2s in EITHER of the factors. Neither one requires three 2s so the multiple doesn't REQUIRE three 2s. There are certainly many common multiples that will include three 2s, but it is not a requirement. When you are looking for a LEAST common multiple you have to use the LEAST possible number of each prime factor.
Got it. Thank you very much.
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by Tani » Sat Jul 09, 2011 6:03 pm
It can be a tricky concept, but is very useful once you get it!
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by JackSparrow » Sat Feb 08, 2014 3:30 pm
Had the HK question said "must 21 be a factor of H+K" instead of "must 21 be a factor of HK", what would the prime box look like then?

Thanks
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