Animated Solution for Mathematics - Statistics: The number of values of $a\in \mathbb{N}$ such that the variance of $3,7,12,a,43-a$ is a natural number is:

A Statistical Journey

Welcome, future engineers! Today, we are going to dissect a problem that looks like a standard statistics question but is actually a beautiful exercise in algebraic symmetry and number theory. We are given a dataset: $3,7,12,a,43-a$.

Our mission is to find how many natural numbers $a$ exist such that the variance of this set is a natural number. Let's embark on this journey.

The Constant Anchor

When you first look at a dataset with a variable, your instinct might be to panic. But take a breath and look at the terms. We have $a$ and $43-a$.

If we calculate the mean, $\bar{x}$, we sum all observations and divide by $n=5$. Watch the magic happen:

Notice how the $+a$ and $-a$ cancel out perfectly? This is the universe telling you that the mean is independent of $a$.

We are left with $\frac{65}{5}$, which gives us a clean, constant mean of $\bar{x}=13$. This is our anchor. No matter what $a$ is, the data will always balance around $13$.

The Variance Beast

Now, we invoke the variance formula: ${\sigma }^2=\frac{\sum x_i^2}{n}-(\bar{x}{)}^2$. We know $n=5$ and $\bar{x}=13$. The real work lies in calculating $\sum x_i^2$.

Let's expand this carefully:

Calculating the constants: $9+49+144=202$. Now, expand $(43-a{)}^2$ using the identity $(x-y{)}^2=x^2-2xy+y^2$. This gives us $1849-86a+a^2$.

Combining everything, we get:

Now, substitute this back into our variance formula:

To combine these, we use a common denominator of $5$. Since $169\times 5=845$, we have:

The Modular Trap

We need ${\sigma }^2$ to be a natural number. This means the numerator $2a^2-86a+1206$ must be divisible by $5$. In the language of number theory, we write this as a congruence:

Let's simplify the coefficients modulo $5$. Since $86=85+1$, $-86\equiv -1(\mathrm{mod}5)$. And $1206=1205+1$, so $1206\equiv 1(\mathrm{mod}5)$.

Our condition becomes:

Now, we test the possible remainders of $a$ when divided by $5$ (the set $\{0,1,2,3,4\}$):

1. If $a\equiv 0$, then $2(0)^2 - 0 + 1 = 1 ot\equiv 0$. 2. If $a\equiv 1$, then $2(1)^2 - 1 + 1 = 2 ot\equiv 0$. 3. If $a\equiv 2$, then $2(4) - 2 + 1 = 7 \equiv 2 ot\equiv 0$. 4. If $a\equiv 3$, then $2(9) - 3 + 1 = 18 - 3 + 1 = 16 \equiv 1 ot\equiv 0$. 5. If $a\equiv 4$, then $2(16) - 4 + 1 = 32 - 4 + 1 = 29 \equiv 4 ot\equiv 0$.

After testing every possible case, we find that the expression $2a^2-a+1$ is never divisible by $5$. It is a mathematical impossibility for the variance to be a natural number for any integer $a$.

Therefore, the number of such values is exactly 0.

This problem teaches us that even when the algebra looks daunting, structure and modular arithmetic can dismantle the complexity. Keep practicing, stay curious, and remember: the math always tells the truth!