If x is a positive integer, is (x)(x + 2)(x + 4) divisible by 12?

Notice that no matter whether x is even or odd, out of x, x+2 and x+4 one must be a multiple of 3. So, we are basically asked to find whether (x)(x + 2)(x + 4) is divisible by 4. Now, if x=odd, then all three multiples are odd, thus (x)(x + 2)(x + 4) will be odd and not divisible by 4. If x=even, then (x)(x + 2)(x + 4)=even*even*even, thus it'll be divisible by 4.

Therefore, the question boils down to find whether x is even.

(1) x^2 + 2x is a multiple of 3 --> if x=1=odd, then the answer is NO but if x=4, then the answer is YES. Not sufficient.

(2) 3x is a multiple of 2. This statement implies that x=even. Sufficient.

Answer: B.

Hope it's clear.
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Merging topics. Please refer to the solution above.
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Here's my interpretation, I'd actually love it if someone could tell me if I'm wrong here as I often make little mistakes in the theory on these questions.

Is (x)(x + 2)(x + 4) divisible by 12?

(1) x^2 + 2x is a multiple of 3.
(2) 3x is a multiple of 2.

I immediately rephrase the 12 into (2^2)*(3). So we need to be divisible by 2 twice and 3 one. So contained in these three terms we need two multiples of 2 and one multiple of 3.

First observation is we either have consecutive even numbers or consecutive odd numbers. If it's the latter, the answer is definitely a 'no' and if it's the former it will be 'yes'.

Why? Here's a quick aside.

By the way this is not something you have time to do in a question so I would really recommend internalizing this concept as it is the backbone of a lot of problems on divisibility, remainders, and several other question families.

Think about the progression of integers (and this is a good thing to get comfortable with). We need two 2's and one 3 from these three integers. In general, we get a 2 every second integer and we get a 3 every third integer.

Now, if we have three odd numbers above what are we going to get? Three numbers that have ZERO 2's. That's what an odd number is: a number that is not divisible by 2. How many 3's will there be? Well we have three consecutive odd numbers, so three numbers that have a range of 4 and start on an odd number.

For instance:

3, 5, 7 or 13, 15, 17 or 21, 23, 25

We can see from these examples that that is always going to be one number that is a multiple of 3. Will it always be only one multiple of 3? Yes, only one. Why? Well, for the same reason we gave above. Multiples of 3 are going to occur every 3rd integer. Perhaps an easier way to understand this (it helps me) is to visualize it.

Consider three consecutive odd numbers:

5, 7, 9

From zero ascending that looks like this:

0 _ _ _ _ 5 _ 7 _ 9 _ ...

Where are the multiples of 3?

0 _ _ 3 _ 5 6 7 _ 9 _

Algebraically, we can say that for three numbers: x, x + 2, and x +4:

If (x) is a multiple of 3 then we have our multiple of three (and we can say the other two integers are not multiples of 3). If x is NOT a multiple of 3 then either (x + 1) or (x - 1) is, but not both. If (x + 1) is, then so is (x + 4) which is the last of our three numbers. If (x + 1) and (x) are not, then (x - 1) is and so is (x + 2) which is our second integer.

So we can see that for any three consecutive odd integers you are going to hit a multiple of 3 at some point in the three numbers.

Therefore in three consecutive odd integers we will have one multiple of 3 and no multiples of 2.

Now, what if we have three even numbers? Well, obviously there a bunch of 2's. What about 3's, how many will there be? Same logic as above. If x is a multiple of 3 we have one there. If x isn't then either (x + 1) or (x - 1) is. If it's the former then x + 4 is a multiple of 3. If it's (x -1) then (x-2) is a multiple of 3.

Ok so before we go to the statements we've proven (albeit laboriously) that the answer depends entirely on whether x is odd or even.

Statement 1: x^2 + 2x is a multiple of 3.

x*(x + 2) multiple of 3.

We don't know if they are even, this could be 6*8 or 3*5. INSUFFICIENT.

Statement 2: 3x is a multiple of 2.

So if 3x is a multiple of 2, then x must be even because 3*(?) = even then the (?) must be even. SUFFICIENT.

Another way to look at statement 2 is to rephrase the statement into:

3x = 2(i), where i is some integer
x = (2/3)*(i)

Therefore for x to remain an integer i must be a multiple of 3. So x is a multiple of 2 and 3. Therefore x is (at a minimum) 6 and we have one 2, one 3 and then our second term in the question will provide the second 2 and we have all we need.

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