by dmitryknowsbest Sat Jan 08, 2011 5:11 pm
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