If n is a positive integer, is n^3 n divisible by 4 ?

Although this question is on finding out divisibility by 4, using the divisibility rule of 4 may not fetch the desired result here. Instead, we need to work on identifying the type of numbers that are part of the given expression.

The given expression, \(n^3\) – n can be broken down into (n-1) * n * (n+1). This is the product of 3 consecutive integers. The product of any 3 consecutive integers is always divisible by 3! i.e. 6.

Note: The product of any n consecutive integers is always divisible by n! and exactly one of these numbers will be exactly divisible by n.


For the product to be divisible by 4, two of the numbers should be even. Any data that helps us establish this will be sufficient data.

From statement I, we know that n = 2k + 1, where k is an integer. This is nothing but the definition of an odd number. So, n is odd; this means that (n-1) and (n+1) is even. Since there are 2 even numbers in the product, the number has to have a four and hence be divisible by 4.

Statement I alone is sufficient. Answer options B, C and E can be ruled out, possible answer options are A or D.

From statement II, \(n^2\) + n is divisible by 6. \(n^2\) + n can be broken down as n*(n+1). This is the product of 2 numbers.
So, as per the statement, the product of 2 numbers is divisible by 6. This means that at least one of these two numbers is even and the other is a multiple of 3. But, we do not know anything about the number (n-1).

For example, if n = 2, (n-1) = 1 and (n+1) = 3. The product of (n-1) * n * (n+1) is not divisible by 4.

On the other hand, if n = 3, (n-1) = 2 and (n+1) = 4. The product of (n-1) * n * (n+1) is divisible by 4.

The data given in statement II is insufficient. Answer option D can be eliminated.
The correct answer option is A.

Hope this helps!