Evaluate the validity of each statement concerning divisibility rules for integers.

We need to evaluate the validity of each statement as a divisibility rule.

• Statement I: An integer $k$ is divisible by 4 if and only if the sum of the digits of $k$ is divisible by 4. This statement is incorrect. For example, consider the number $k=12$. The sum of its digits is $1+2=3$, which is not divisible by 4. However, 12 is divisible by 4. The correct rule for divisibility by 4 is that the number formed by its last two digits must be divisible by 4.

• Statement II: An integer $k$ is divisible by 10 if and only if the number formed by the last 3 digits of $k$ is divisible by 10. This statement is correct. If an integer $k$ is divisible by 10, its last digit must be 0. If the last digit is 0, then the number formed by its last 3 digits (e.g., $d_2 d_1 0$) will also end in 0, and thus be divisible by 10. Conversely, if the number formed by the last 3 digits of $k$ is divisible by 10, it implies that the last digit of $k$ must be 0. If the last digit of $k$ is 0, then $k$ itself is divisible by 10.

• Statement III: An integer $k$ is divisible by 3 if and only if the number formed by the last 2 digits is divisible by 3. This statement is incorrect. For example, consider the number $k=123$. The number formed by its last two digits is 23, which is not divisible by 3. However, 123 is divisible by 3 (since the sum of its digits $1+2+3=6$ is divisible by 3). The correct rule for divisibility by 3 is that the sum of its digits must be divisible by 3.

Only statement II is a valid way to verify divisibility by 10. Therefore, none of the options A, B, C, or D are correct, as they all include I or III.

The correct option is E.

$$\boxed{\text{E}}$$

We are given $a=24$ and $b=36$. We need to find the greatest common divisor (GCD) of $a$ and $b$.

First, find the prime factorization of each number:

$$24 = 2 \times 12 = 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1$$ $$36 = 2 \times 18 = 2 \times 2 \times 9 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2$$

To find the GCD, we take the lowest power of each common prime factor:

$$\gcd(24, 36) = 2^{\min(3,2)} \times 3^{\min(1,2)}$$ $$\gcd(24, 36) = 2^2 \times 3^1$$ $$\gcd(24, 36) = 4 \times 3$$ $$\gcd(24, 36) = 12$$

The greatest common divisor of 24 and 36 is 12.

The correct option is C.

$$\boxed{\text{C}}$$

The question asks about the base case typically used when proving a statement by mathematical induction.

Mathematical induction is a method used to prove that a statement $P(n)$ is true for all natural numbers $n$ (or for all $n$ greater than or equal to some initial integer). It consists of two steps:

1. Base Case: Prove that the statement $P(n)$ is true for the initial value of $n$ (e.g., $n=1$ or $n=0$).
2. Inductive Step: Assume that $P(k)$ is true for an arbitrary natural number $k \ge n_0$ (the inductive hypothesis), and then prove that $P(k+1)$ is also true.

Let's evaluate the given options:

• A. The statement is assumed to be true for all $n \in \mathbb{N}$. This is incorrect; we aim to prove it for all $n$, not assume it.
• B. The statement is true for the smallest value of $n$. This describes the base case, but option C is more complete.
• C. The statement is proven true for the initial value, usually $n=1$ or $n=0$. This accurately describes the base case, emphasizing that it must be proven for a specific initial value.
• D. The statement is assumed true for $n+1$. This is part of the inductive step, where we assume $P(k)$ is true to prove $P(k+1)$.

Option C provides the most accurate and complete description of the base case in mathematical induction.

The correct option is C.

$$\boxed{\text{C}}$$