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naifsds
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Post subject: divisbility  Posted: Mon Jan 03, 2011 9:45 pm |
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If x is divisible by 144, and cubed root to the x is an integer (sorry was not able to find the root symbol.
Which of the following is "cubed root to the x" definitely divisible by? - choose all that apply
4
8
9
12
(i got this question from the manhattan gmat flash card app)
would it be possible to mention other examples of divisibility, find this subject a little confusing
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naifsds
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Post subject: Re: divisbility Posted: Thu Jan 06, 2011 11:02 pm |
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can some one reply please
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emailnaik
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Post subject: Re: divisbility Posted: Fri Jan 07, 2011 1:57 pm |
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144 = 2*2*2*2*3*3
cube root of x is an integer therefore the smallest number which happens to be a cube of x and divisible by 144 is 1728=144*12
1728 = 2*2*2*2*2*2*3*3*3
Given the constraints if the cube of x is divisible by 144 then that number will be divisible by 4, 8, 9, and 12.
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dmitryknowsbest
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Post subject: Re: divisbility Posted: Sat Jan 08, 2011 5:11 pm |
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Thanks for the response, emailnaik. You can also use the prime box here. Since x is a multiple of 144, it must have 2*2*2*2*3*3 in its prime box.
We also know that x is the cube of an integer. For this to be true, all of its prime factors must appear in groups of three. We can see this if we look at some actual cubed integers. To find the cube root, we just separate the primes into three groups:
8 = 2x2x2. The cube root is 2.
216=2x2x2x3x3x3 = (2x3)(2x3)(2x3). The cube root is 2x3=6.
We can tell that 144 is not a cubed integer, since we end up with leftover primes:
144 = (2x3)(2x3)(2x2). The groups are not all the same, so we don't have a cubed integer.
So how could we make a cubed integer with these primes? We currently have four 2's and two 3's. To make equal groups, we need to multiply in at least two more 2's and one more 3. Then we will have the following:
2*2*2*2*2*2*3*3*3= (2*2*3)(2*2*3)(2*2*3). The cube root of x is 2*2*3=12. Looking at the primes in x, we can easily construct 4, 8, 9, or 12, so x is divisible by all of them.
_________________
Dmitry Farber
Manhattan GMAT Instructor
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dmitryknowsbest
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Post subject: Re: divisbility Posted: Sat Jan 08, 2011 5:16 pm |
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Oh, and for additional problems, check out Part II of the Number Properties strategy guide, and then follow up with the Advanced OG assignments in divisibility & primes.
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Dmitry Farber
Manhattan GMAT Instructor
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drew.eng
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Post subject: Re: divisbility Posted: Mon Jan 10, 2011 3:28 am |
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Wait...we have to consider the factors of 12, and so the correct answer should be a)4 and d)12. 12 (2*2*3) is not divisible by 8 or 9.
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dmitryknowsbest
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Post subject: Re: divisbility Posted: Mon Jan 10, 2011 3:50 am |
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Ah, yes, sorry. I didn't look back at the original question. (Common error to watch for on the test!) x must be divisible by all the numbers listed, but the cube root of x only needs to be divisible by the factors of 12. The only factors of 12 listed are 4 and 12.
Good call!
_________________
Dmitry Farber
Manhattan GMAT Instructor
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