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If J is divisible by 12 and 10, is J divisible by 24? The [#permalink]
 22 Apr 2007, 22:32
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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VP
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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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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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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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Director
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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.
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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.
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