Confusing MGMAT quant problem on Number Properties

Question

HI everyone. I'm very very very new to this forum. I am studying for the GMAT and started with the MGMAT books. I'm doing Number Properties right now. There's a question in their first chapter that goes like this:

Is x divisible by 120?
(1) x is divisible by 12
(2) x is divisible by 30

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.

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.

My question is, if a 2 is considered redundant (as the MGMAT answer key reminds you), then why wouldn't this apply to other numbers that appear in both prime boxes as well, i.e., the number 3 appears as a factor of both 12 and 30.

Thank you very much for your time. 
Dors

pike · Answer

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...

arunangsude2011 · Answer

good explanation, regards

Raghava747 · Answer

Let us take A. first : 
divisible by 12: numbers divisible by 12 are 12,24,36,48...... 120.....

hence A alone not sufficient. since we have numbers less than 120 divisible by 12

Let us take B. 
divisible by 30: numbers divisible by 30 are 30,60,90,120....

Hence B alone is not sufficient. Since we have numbers less than 120 divisible by 30

Let us combine : We have number divisible by 12 and 30 :
to find combined divisibility we take LCM. LCM of 12 and 30 can be found by reducing each number to its prime factors

12 : 2*3*2
30: 2*3*5

Now LCM can be obtained by taking highest powers of each prime factor and multiplying them 2^2*3^1*5^1 i.e 2*2*3*5 = 60

This number is also not divisible by 120.

Hence option E. We need more data and both together are in sufficient.

dorsvenabili · Answer

Dear pike and raghava747: THANK YOU SO MUCH for your explanations. I am SO SORRY it took me so long to respond, but I'm living in an island in the Caribbean with haphazard electricity AND spotty internet service. I finally understood that the 3 could have been seen twice and hence; the option is E. Thank you very very much again. I really really appreciate it.

Lorenzo3688 · Answer

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

x = 60 = 2*3*5*2

ccooley · Answer

A good way to get your head around this sort of problem is to think of the two divisors as "pieces of evidence". You have one piece of evidence that says the number can be divisible by 12, which tells you that it has at least one three and at least two twos in its prime factorization. The other piece of evidence tells you that it's divisible by 30, so it has at least one three, at least one two, and at least one five in its prime factorization. If you knew both of those facts:

"The number has at least one 3 and at least two 2s."
"The number has at least one 3, at least one 2, and at least one 5."

Then what could you say about the number? You definitely know that it has at least one 3; both pieces of evidence tell you that. You also know that it has at least two 2s, and in that case, the first piece of evidence gives you more information than the second one, so you can ignore the second one. Same with the 5s: the second piece of evidence gives you new information that you didn't get from the first one, so you can use it.

Combining the pieces, then, you get:

"The number has at least two 2s, at least one 3, and at least one 5."

Or in other words, it's divisible by 60 and who knows what else.

Confusing MGMAT quant problem on Number Properties

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