dorsvenabili wrote:
So I drew the prime boxes for 12 and 30, and obtained, correctly that 12's prime factors are 2,2,3 and 30's prime factors are 2, 3, 5. Then, I constructed the prime box for x as 2, 2, 5.... Notice that I did not include another 2 because it is redundant. I applied the same logic to not including the 3, since it appeared in both prime boxes. I did get the right answer, i.e., (E) cannot be determined, but apparently for the wrong reasons, which frighten me.
According to the MGMAT answer key, the prime box for x SHOULD contain the 3. In other words, the prime box for x should be: 2, 2, 3, 5.
It is not redundant. Think of it like this.
Is x divisible by 4?
x = 2 <---- no
x = 2*2 <---- yes
Clearly we need to know the
number of twos to answer this question.
To know that x is divisible by 120, it
must have 2,2,2,3,5 as its prime factors (at a minimum). If it has only one 2, then it will not be divisible by 120.
Now on the other hand, we are told through 1) and 2) that x is divisible by 12 and 30. The most information we can glean from this is that
x = 2,2,3,5
Note that when we do THIS, we don't just lump all factors from 1) and 2) and get
x = 2,2,2,3,3,5
This is because we could have 'seen' the some number twice. Consider
x = 60 = 2*3*5*2
This is divisible by 1) and 2), yet is not divisible by 120...
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