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If J is divisible by 12 and 10, is J divisible by 24? The

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CEO
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If J is divisible by 12 and 10, is J divisible by 24? The [#permalink] New post 22 Apr 2007, 21:32
If J is divisible by 12 and 10, is J divisible by 24?

The official ManhattanGMAT answer: cannot be determined.

"If J is divisible by 12 and by 10, its prime factors are 2, 2, 3 and 5. Therefore, any integer that can be constructed as a produt of any of these integers is also a factor of J.

24= 2 x 2 x 2 x 3. There are only two 2's in the prime box; therefore, 24 is not necessarily a factor"



my question is, is this an error?

12 and 10 yields factors of 2,2,3,5,2. There are THREE 2s. Hence 24 is a factor.
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Re: factors [#permalink] New post 22 Apr 2007, 21:45
bmwhype2 wrote:
If J is divisible by 12 and 10, is J divisible by 24?

The official ManhattanGMAT answer: cannot be determined.

"If J is divisible by 12 and by 10, its prime factors are 2, 2, 3 and 5. Therefore, any integer that can be constructed as a produt of any of these integers is also a factor of J.

24= 2 x 2 x 2 x 3. There are only two 2's in the prime box; therefore, 24 is not necessarily a factor"



my question is, is this an error?

12 and 10 yields factors of 2,2,3,5,2. There are THREE 2s. Hence 24 is a factor.


I don't think its a mistake

you have to look at 10 & 12 separately 10 = 2*5 and 12 = 3*2*2
together they have 2,2,2,3,5 but one of the 2 is repetitive number.

for example:

60 = 5*2*2*3

60/12 = 5

60/10 = 6

but 60/24 = 2.5 - all because the of the mutual 2 !

:)
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Re: factors [#permalink] New post 22 Apr 2007, 22:13
KillerSquirrel wrote:
bmwhype2 wrote:
If J is divisible by 12 and 10, is J divisible by 24?

The official ManhattanGMAT answer: cannot be determined.

"If J is divisible by 12 and by 10, its prime factors are 2, 2, 3 and 5. Therefore, any integer that can be constructed as a produt of any of these integers is also a factor of J.

24= 2 x 2 x 2 x 3. There are only two 2's in the prime box; therefore, 24 is not necessarily a factor"



my question is, is this an error?

12 and 10 yields factors of 2,2,3,5,2. There are THREE 2s. Hence 24 is a factor.


I don't think its a mistake

you have to look at 10 & 12 separately 10 = 2*5 and 12 = 3*2*2
together they have 2,2,2,3,5 but one of the 2 is repetitive number.

for example:

60 = 5*2*2*3

60/12 = 5

60/10 = 6

but 60/24 = 2.5 - all because the of the mutual 2 !

:)


thanks, i really appreciate the explanation.
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 [#permalink] New post 22 Apr 2007, 22:13
If X is div by 12 then: X / (2^2 * 3) is an integer
If X is div by 10 then: X / (2 * 5) is an integer

The conclusion you can make is that X can be divided by 2^2 * 3 * 5 and still be an integer. If you factorize the number 24 = 2^3 * 3 you will notice the presence of 2^3 , so based on the conclusion drawn before you can not establish that X will be also divisible by 24.
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Re: factors [#permalink] New post 22 Apr 2007, 23:58
bmwhype2 wrote:
If J is divisible by 12 and 10, is J divisible by 24?



12 = 2^2*3
10 =2*5

Smallest positive common multuple of these two numbers is 2^2*3*5=60
So, J can be 60, 120 and so on...

If J=60 ===> J/24 IS NOT integer
IF J=120 ===> J/24 = 5 - i.e. IS integer
Therefore, the answer is Yes and NO - i.e. can not be determined.
CEO
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Re: factors [#permalink] New post 13 May 2007, 08:51
botirvoy wrote:
bmwhype2 wrote:
If J is divisible by 12 and 10, is J divisible by 24?



12 = 2^2*3
10 =2*5

Smallest positive common multuple of these two numbers is 2^2*3*5=60
So, J can be 60, 120 and so on...

If J=60 ===> J/24 IS NOT integer
IF J=120 ===> J/24 = 5 - i.e. IS integer
Therefore, the answer is Yes and NO - i.e. can not be determined.


thanks. much easier to understand now.
Re: factors   [#permalink] 13 May 2007, 08:51
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If J is divisible by 12 and 10, is J divisible by 24? The

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