Classify the following numbers as prime or composite: 201, 211, 55, 112, 73. Then, find the product of odd numbers between 27 and 32, and determine the difference between the count of even and odd dig

An irrational number cannot be expressed as a simple fraction $\frac{p}{q}$ and has a non-terminating, non-repeating decimal expansion. Square roots of non-perfect squares are irrational.

• $\sqrt{79}$: Since 79 is not a perfect square ($8^2=64$, $9^2=81$), $\sqrt{79}$ is an irrational number.
• $\sqrt{2.197}$: This can be written as $\sqrt{\frac{2197}{1000}}$. Since 2197 is not a perfect square (it ends in 7), $\sqrt{2197}$ is irrational. Therefore, $\sqrt{2.197}$ is an irrational number.
• $\sqrt{30.42}$: This can be written as $\sqrt{\frac{3042}{100}}$. Since 3042 is not a perfect square (it ends in 2), $\sqrt{3042}$ is irrational. Therefore, $\sqrt{30.42}$ is an irrational number.

The irrational numbers from the list are $\boxed{\sqrt{79}, \sqrt{2.197}, \sqrt{30.42}}$.

This expression is in the form of a difference of squares, $a^2 - b^2 = (a-b)(a+b)$.
Here, $a^2 = 81 \implies a = 9$.
And $b^2 = (q-7)^2 \implies b = (q-7)$.

Substitute these into the difference of squares formula:

$$81 - (q-7)^2 = [9 - (q-7)][9 + (q-7)]$$ $$= (9 - q + 7)(9 + q - 7)$$ $$= (16 - q)(2 + q)$$

The factored form is $\boxed{(16 - q)(2 + q)}$.

First, find the prime factorization of 729:

$$729 \div 3 = 243$$ $$243 \div 3 = 81$$ $$81 \div 3 = 27$$ $$27 \div 3 = 9$$ $$9 \div 3 = 3$$ $$3 \div 3 = 1$$

So, $729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6$.

Next, find the prime factorization of 2187:

$$2187 \div 3 = 729$$

Since $729 = 3^6$, we have:

$$2187 = 3 \times 3^6 = 3^7$$

Now, evaluate $\frac{729}{2187}$:

$$\frac{729}{2187} = \frac{3^6}{3^7}$$

Using the exponent rule $\frac{a^m}{a^n} = a^{m-n}$:

$$\frac{3^6}{3^7} = 3^{6-7} = 3^{-1} = \frac{1}{3}$$

The index forms are $729 = \boxed{3^6}$ and $2187 = \boxed{3^7}$.
The evaluation is $\frac{729}{2187} = \boxed{\frac{1}{3}}$.