If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5

Answer: (B) -2.

The explanation says to test the numbers and will find that choice (B) -2, will not result in an integer. However, looking at how it is done, answer choice (D) 2 & (E) 5 also doesn't work because it can sometimes result in ZERO which is not divisible by 3, or is it? Sufficient to say that then it is ZERO so it is divisible by 3?

Zero is divisible by anything. Test it out: divisibility means that, when you divide by some number, your answer is an integer with no remainder. Divide 0 by 3. What do you get? 0, with no remainder. That passes the divisibility test.

Hi,
Another way of looking at this is as follows.
Product of three consecutive integers will always be divisible by 3. Thus, (x-1)x(x+1) is always divisible by 3.
So k = -1 resuslts in the product being divisible by 3. So if k=-1 is divisible then k = -4 (-1 - 3), k = 2 (-1 + 3) and k = 5 (-1 + 6) will always be divisible by 3.
That just leaves out -2. Which is the correct answer

Thanks, Captain. (And just to make something explicit: Captain came up with k=-1 first because that is what k would have to be to get the algebraic representation of three consecutive integers: (x-1)x(x+1). That last term is the x-k term, so if k=-1, then the term becomes x+1.)

Guys, what am I missing here ?

We are being asked to find a value of k that makes the expression x(x-1)(x-k) not divisible by 3. This means that if any one of the parts become divisible by 3, the entire expression is divisible by 3.

We have already seen that only -2 is an exception hence the answer. But if k were -2, exp x-k would become x+2. Then if x=4 or 7, x+2 becomes 6 and 9 respectively, which are divisible by 3.

I must be missing the part that prevents x from being 4 or 7. Please advise.

I have now seen explanation of the same question in an another post. Although not convinced of the question itself, I do not need further explanation. Thanks.

Replying to viksnme's post:

The key word in the question is "must." You presented specific values for x such that k=-2 does result in a number divisible by 3. But you cannot guarantee that for ALL values of x, the entire expression will be divisible by 3 if k=-2. Try plugging in k=-2 and x = 5 for a counterexample.

x(x – 1)(x – k)
5(4)(7)
not divisible by 3

Rey