If a^3 - a = b, and b is even, then b must be divisible by what number

Question

If  a^3 - a = b , where a is an integer and  b  is even, then b must be divisible by what number?
 
(A) 1
(B) 4
(C) 6
(D) 8
(E) 10

Abhishek009 · Answer

Plug in the least even value of a = 2 ; So, a^3 - a = 6
Plug in the least odd value of a = 3 ; So, a^3 - a = 24

Thus, b must be divisible by 6 , Answer must be (C)

GauriK · Answer

a^3 - a = b then a(a^2- 1) = b
which is a * (a+1) *(a-1) 
These are three consecutive integers. 
# Rule: Three consecutive integers are always divide by 2 and 3. As 2 & 3 are not in option. Choice C i.e. 6 is correct answer.

Ozzy11100 · Answer

I understand what the question is asking, and thus why the prescribed answer is C. But of course, A is also a correct answer.

Any integer B MUST also be divisible by 1.

I don't think there should be an answer choice 1.

GMATWhizTeam · Answer

Solution:

There are a couple of problems with this question:

1. Option A is 1. Every number is divisible by that. So will  b  no matter itsv value
ature.

2.  b  is even. This seems to be redundant information.

Bunuel  please look into these.

Now, we have:  a^3 - a = b

= a(a^2-1)=b

= a(a^2-1^1)=b

= a(a+1)(a-1)=b

= (a-1)a(a+1)=b

Since a is an integer,  a-1, a  and  a+1  will be 3 consecutive integers. We know that  product of 3 consecutive integers is always divisible by 6 . We should know this as a property.

Hence the right answer is    Option C   .

If a^3 - a = b, and b is even, then b must be divisible by what number

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