If x is a whole number, then the expression x²(x² − 1) is always divisible by which of the following numbers?

Introduction / Context:
This problem tests your understanding of divisibility properties and factorisation for algebraic expressions. You must analyse x²(x² − 1) for whole number values of x and determine by which number it is always divisible. Instead of checking individual values one by one, the key is to factor the expression and use properties of consecutive integers, parity and divisibility by 3. This is a classic number theory style question in aptitude exams.

Given Data / Assumptions:

• x is a whole number (0, 1, 2, 3, and so on).
• Expression: x²(x² − 1).
• We must decide whether it is always divisible by 4, 8, 12, 16 or by none of these.
• Whole numbers include 0, and 0 is divisible by any positive integer.

Concept / Approach:
Start by factoring the expression algebraically. Note that x² − 1 is a difference of squares, so x² − 1 = (x − 1)(x + 1). Thus x²(x² − 1) = x²(x − 1)(x + 1). This product contains the three consecutive integers x − 1, x and x + 1, and one extra factor of x. The product of three consecutive integers is always divisible by 6 because among any three consecutive integers there is at least one multiple of 2 and one multiple of 3. The extra factor of x ensures additional powers of 2 in many cases. We will check divisibility by 12, 8 and 4 in turn.

Step-by-Step Solution:
Write x²(x² − 1) as x²(x − 1)(x + 1). Observe that (x − 1), x and (x + 1) are three consecutive integers. Among three consecutive integers, one is always a multiple of 3, so the product is divisible by 3. Also, among these three integers at least one is even, so the product is divisible by 2. Therefore x(x − 1)(x + 1) is always divisible by 6. Now consider divisibility by 4. If x is even, then x² has a factor of 4 so the whole product is divisible by 4. If x is odd, then both (x − 1) and (x + 1) are even and consecutive multiples of 2, so one of them is a multiple of 4. Thus the product x²(x − 1)(x + 1) is always divisible by 4 in either case. Since the expression is always divisible by both 3 and 4, it is always divisible by lcm(3, 4) = 12. However it is not always divisible by 8; for example when x = 2 the value is 4 * 3 = 12, which is not a multiple of 8.

Verification / Alternative check:
Test a few small whole numbers to confirm. For x = 0 the expression is 0, divisible by 12. For x = 1 we get 1²(1² − 1) = 1 * 0 = 0, again divisible by 12. For x = 2 we get 4 * 3 = 12, exactly a multiple of 12. For x = 3 we get 9 * 8 = 72, and 72 / 12 = 6. For x = 4 we get 16 * 15 = 240, and 240 / 12 = 20. These checks support our general reasoning.

Why Other Options Are Wrong:

• 4: The expression is indeed always divisible by 4, but this is not the largest guaranteed divisor among the options.
• 8: For x = 2 the value is 12, which is not divisible by 8, so 8 fails.
• 16: For x = 2 or x = 3 the resulting values are not multiples of 16.
• None of these: This is incorrect because we have shown that 12 always divides x²(x² − 1).

Common Pitfalls:
A common error is to test only one or two values of x and then guess a pattern without algebraic justification. Another mistake is to overlook the factorisation x² − 1 = (x − 1)(x + 1) and therefore miss the property of three consecutive integers. For divisibility questions, always look for factorisations, parity patterns and use of least common multiples, rather than relying on trial alone.

Final Answer:
For any whole number x, the expression x²(x² − 1) is always divisible by 12.