Question

For the real variable \(x\), show that a sufficient, but not necessary, condition for \(f(x)=x(x+1)(2 x+1)\) to be divisible by 6 is that \(x\) is an integer.

Short answer

If x is an integer, f(x) is divisible by 6.

Step-by-step solution

Express the function

Given the function, express it as follows: f(x) = x (x + 1) (2x + 1)

Check for integer values

Consider the expression in the context of an integer value for x. When x is an integer, both (x + 1) and (2x + 1) are also integers. Therefore, we have the product of three consecutive integers.

Analyze divisibility by 2

When x is an integer, either x or (x + 1) is always even. Thus, the product will include a factor of 2.

Analyze divisibility by 3

When x is an integer, one of the three consecutive integers (x, x+1, or 2x+1) must be divisible by 3. This guarantees that the product includes a factor of 3.

Conclusion on divisibility by 6

Since the product of the three factors includes both a factor of 2 and a factor of 3, if x is an integer, the function f(x) is divisible by 6.

Prove sufficiency but not necessity

While x being an integer ensures divisibility by 6, it is not the only possible case. For example, for half-integer values like x = 0.5, f(x) = 0, which is divisible by 6, showing that being an integer is sufficient for divisibility but not necessary.

Key concepts

Divisibility

Divisibility is a fundamental concept in mathematics, which helps us determine if one number can be evenly divided by another without leaving a remainder. In our exercise, we want to show that the function \(f(x) = x(x+1)(2x+1)\) is divisible by 6 under certain conditions. For a number to be divisible by 6, it must be divisible by both 2 and 3. This is because 6 is the product of 2 and 3. When we write a number as a product of other integers, one or more of those numbers might have factors of 2 or 3. If, together, those products cover both factors, our number is divisible by 6.

Integer Properties

Integers are whole numbers that can be positive, negative, or zero. Understanding the properties of integers is vital in this problem. When we consider integer values for \(x\), it heavily influences our function \(f(x) = x(x+1)(2x+1)\).

Here are some key properties of integers:

• Adjacent integers: If \(x\) is an integer, then \(x+1\) and \(x-1\) are also integers.
• Parity: An integer can either be even or odd.
• Divisibility: Some integers are divisible by other integers (e.g., 12 is divisible by 2, 3, and 6).

For our function, when \(x\) is an integer, both \(x+1\) and \(2x+1\) will also be integers, making it easier to analyze whether the product is divisible by 6.

Product of Integers

When we form the product of several integers, the resulting value retains factors from each multiplicand. In this case, \(f(x) = x(x+1)(2x+1)\) is the product of three expressions. To determine the divisibility by 6, we can break it down:

• Divisibility by 2: One of \(x\) or \(x+1\) will always be even if \(x\) is an integer.
• Divisibility by 3: Among \(x, x+1, \text{and} 2x+1\), at least one must be divisible by 3 when \(x\) is an integer.

This means that the product \(x(x+1)(2x+1)\) always includes a multiplication by both 2 and 3 when \(x\) is an integer. Hence, it is divisible by 6.

Sufficient Conditions

A sufficient condition guarantees that a certain outcome is achievable under that condition. In this problem, we established that \(f(x) = x(x+1)(2x+1)\) is divisible by 6 if \(x\) is an integer. This is because any integer \(x\) makes at least one factor even (ensuring divisibility by 2) and at least one factor divisible by 3 (ensuring divisibility by 3).

However, this does not mean being an integer is the only way. For example, if \(x = 0.5\), \(f(x) = 0\), which is divisible by 6. Therefore, integer \(x\) is sufficient for divisibility by 6 but not strictly necessary. There are other instances (like half-integers) where \(f(x)\) still satisfies the divisibility requirement.