1. Key Concepts and Formulas

To solve this problem, we will utilize fundamental statistical definitions and a key concept from number theory:

• Natural Numbers ($\mathbb{N}$): The set of positive integers $\{1, 2, 3, \dots\}$. Both the input value 'a' and the resulting variance must belong to this set.
• Mean ($\overline{x}$): For a data set $x_1, x_2, \dots, x_n$, the mean is the sum of all observations divided by the number of observations: $\overline{x} = \frac{\sum_{i=1}^n x_i}{n}$
• Variance ($\sigma^2$): A measure of how spread out the data points are from the mean. A computationally efficient formula for variance is: $\sigma^2 = \frac{\sum_{i=1}^n x_i^2}{n} - (\overline{x})^2$ This formula is particularly useful when the mean is a simple value or when the sum of squares is easy to compute.
• Modular Arithmetic: A system of arithmetic for integers, where numbers "wrap around" when reaching a certain value—the modulus. It's crucial for efficiently checking divisibility conditions without testing infinite values. If an expression $E(a)$ must be divisible by $k$, then $E(a) \equiv 0 \pmod k$.

2. Step-by-Step Solution

We are given five observations: $x_1 = 3, x_2 = 7, x_3 = 12, x_4 = a, x_5 = 43 - a$. The number of observations is $n=5$.

Step 1: Calculate the Mean ($\overline{x}$)

Our first step is to find the mean of the given data set. The mean is a foundational component for calculating variance. $\overline{x} = \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5}$ Substitute the given observations: $\overline{x} = \frac{3 + 7 + 12 + a + (43 - a)}{5}$ Reasoning: Notice how the terms involving 'a' are designed to cancel out. This simplification is a common technique in competitive exams to streamline calculations and focus the problem on subsequent algebraic or number theory challenges. $\overline{x} = \frac{(3 + 7 + 12 + 43) + (a - a)}{5}$ $\overline{x} = \frac{65 + 0}{5}$ $\overline{x} = \frac{65}{5}$ $\overline{x} = 13$ The mean of the data set is $13$, which is a constant and independent of the value of $a$.

Step 2: Calculate the Sum of Squares ($\sum x_i^2$)

Next, we need to compute the sum of the squares of each observation, $\sum x_i^2$, as required by our chosen variance formula. $\sum x_i^2 = 3^2 + 7^2 + 12^2 + a^2 + (43 - a)^2$ Reasoning: This term is a direct input for the variance formula. Careful expansion of $(43-a)^2$ is critical to avoid errors. $\sum x_i^2 = 9 + 49 + 144 + a^2 + (43^2 - 2 \cdot 43 \cdot a + a^2)$ Calculate $43^2 = 1849$. $\sum x_i^2 = 9 + 49 + 144 + a^2 + (1849 - 86a + a^2)$ Combine the constant terms, terms with $a$, and terms with $a^2$: $\sum x_i^2 = (9 + 49 + 144 + 1849) + (a^2 + a^2) - 86a$ $\sum x_i^2 = 2051 + 2a^2 - 86a$ Rearranging in standard quadratic form: $\sum x_i^2 = 2a^2 - 86a + 2051$

Step 3: Calculate the Variance ($\sigma^2$)

Now, we substitute the calculated mean ($\overline{x} = 13$) and sum of squares ($\sum x_i^2 = 2a^2 - 86a + 2051$) into the variance formula: $\sigma^2 = \frac{\sum x_i^2}{n} - (\overline{x})^2$ $\sigma^2 = \frac{2a^2 - 86a + 2051}{5} - (13)^2$ Reasoning: This step combines our previous calculations to express the variance as a function of 'a'. To simplify, we'll combine the fractional and integer terms. $\sigma^2 = \frac{2a^2 - 86a + 2051}{5} - 169$ To combine these, find a common denominator: $\sigma^2 = \frac{2a^2 - 86a + 2051 - (169 \times 5)}{5}$ Calculate $169 \times 5 = 845$. $\sigma^2 = \frac{2a^2 - 86a + 2051 - 845}{5}$ $\sigma^2 = \frac{2a^2 - 86a + 1206}{5}$ This is the complete expression for the variance in terms of $a$.

Step 4: Apply the Condition for Variance to be a Natural Number

We are given that $a \in \mathbb{N}$ (a natural number) and we need the variance $\sigma^2$ to also be a natural number ($\sigma^2 \in \mathbb{N}$). For $\sigma^2$ to be a natural number, two conditions must be met:

1. $\sigma^2$ must be an integer.
2. $\sigma^2$ must be positive (natural numbers are positive integers).

Let's check the positivity condition first. The numerator is a quadratic $P(a) = 2a^2 - 86a + 1206$. The discriminant of this quadratic is $D = b^2 - 4ac = (-86)^2 - 4(2)(1206) = 7396 - 9648 = -2252$. Reasoning: Since the discriminant $D < 0$ and the leading coefficient (2) is positive, the quadratic $P(a)$ is always positive for all real values of $a$. Therefore, $\sigma^2 = \frac{P(a)}{5}$ will always be positive. If it's an integer, it will automatically be a natural number.

The crucial condition then simplifies to $\sigma^2$ being an integer. This implies that the numerator, $2a^2 - 86a + 1206$, must be perfectly divisible by 5.

Step 5: Analyze Divisibility by 5 using Modular Arithmetic

For $2a^2 - 86a + 1206$ to be divisible by 5, its value modulo 5 must be 0. We will use modular arithmetic to simplify the expression and test possible values for $a \pmod 5$. $2a^2 - 86a + 1206 \equiv 0 \pmod 5$ Let's find the residues of the coefficients modulo 5:

• $2 \pmod 5 \equiv 2$
• $-86 \pmod 5$: Since $86 = 17 \times 5 + 1$, $86 \equiv 1 \pmod 5$. Therefore, $-86 \equiv -1 \pmod 5$.
• $1206 \pmod 5$: Since $1206 = 241 \times 5 + 1$, $1206 \equiv 1 \pmod 5$.

Substituting these into the congruence: $2a^2 - a + 1 \equiv 0 \pmod 5$ Since $a \in \mathbb{N}$, $a$ can take any positive integer value. We only need to test the possible residues of $a$ modulo 5 (i.e., $0, 1, 2, 3, 4$), as the pattern of $a^2$ and $a$ modulo 5 repeats every 5 values.

• If $a \equiv 0 \pmod 5$: $2(0)^2 - 0 + 1 = 1 \not\equiv 0 \pmod 5$
• If $a \equiv 1 \pmod 5$: $2(1)^2 - 1 + 1 = 2 - 1 + 1 = 2 \not\equiv 0 \pmod 5$
• If $a \equiv 2 \pmod 5$: $2(2)^2 - 2 + 1 = 2(4) - 2 + 1 = 8 - 2 + 1 = 7 \equiv 2 \pmod 5 \not\equiv 0 \pmod 5$
• If $a \equiv 3 \pmod 5$: $2(3)^2 - 3 + 1 = 2(9) - 3 + 1 = 18 - 3 + 1 = 16 \equiv 1 \pmod 5 \not\equiv 0 \pmod 5$
• If $a \equiv 4 \pmod 5$: $2(4)^2 - 4 + 1 = 2(16) - 4 + 1 = 32 - 4 + 1 = 29 \equiv 4 \pmod 5 \not\equiv 0 \pmod 5$

In every possible case for $a \pmod 5$, the expression $2a^2 - a + 1$ does not evaluate to $0 \pmod 5$. This means that for no natural number $a$ is the numerator $2a^2 - 86a + 1206$ divisible by 5.

Conclusion: Since the numerator is never divisible by 5 for any natural number $a$, the variance $\sigma^2 = \frac{2a^2 - 86a + 1206}{5}$ will never be an integer. Consequently, it can never be a natural number. Therefore, there are no values of $a \in \mathbb{N}$ that satisfy the given condition.

3. Common Mistakes & Tips

• Algebraic Errors: Be meticulous when expanding terms like $(43-a)^2$. A common mistake is to forget the middle term ($2ab$) in $(a-b)^2$.
• Arithmetic Errors: Double-check all numerical calculations, especially multiplication and subtraction, as small errors can propagate.
• Definition of Natural Numbers: Remember that natural numbers are positive integers (1, 2, 3, ...). The variance must be both an integer and greater than zero.
• Modular Arithmetic Application: When simplifying expressions modulo $k$, ensure all coefficients and constants are reduced modulo $k$ before testing values of $a$. This makes the testing process much faster and covers all integer possibilities.
• Discriminant Interpretation: For a quadratic $Ax^2+Bx+C$, if $A>0$ and the discriminant $D<0$, the quadratic is always positive. This is a useful shortcut to confirm the positivity condition for variance without further analysis of the quadratic roots.

4. Summary

This problem required us to first calculate the mean and then the variance of a given data set, expressing the variance as a function of 'a'. We found the mean to be $13$ and the variance to be $\sigma^2 = \frac{2a^2 - 86a + 1206}{5}$. For the variance to be a natural number, it must be a positive integer. We confirmed its positivity using the discriminant of the numerator's quadratic. The core of the problem then became finding values of natural number 'a' for which the numerator is divisible by 5. By employing modular arithmetic and testing all possible residues of 'a' modulo 5, we demonstrated that the numerator is never divisible by 5. Thus, the variance can never be an integer, and consequently, never a natural number.

5. Final Answer

The number of values of $a \in \mathbb{N}$ such that the variance is a natural number is 0.

The final answer is $\boxed{\text{0}}$, which corresponds to option (A).