Prime Factorization: Recognizing divisibility rules

To solve this problem, let's analyze the numbers:

• Step 1: Identify that all numbers of the form $?4$ end with the digit 4, which is even.
• Step 2: Apply the divisibility rule for 2: A number ending in an even digit is divisible by 2.
• Step 3: Recognize that since all such numbers are even due to ending in 4, each number is divisible by 2.

Therefore, every number ending in 4 has 2 as a factor.

To further clarify, consider some examples: 14, 24, 34, ..., 94. These can all be divided by 2.

Therefore, the prime factor that will surely appear among the factors of any number in the form $"?4"$ is $2$.

To solve this problem, follow these steps:

• Step 1: Recognize that a two-digit number ending in 0 can be expressed as $10 \times n$, where $n$ is an integer from 1 to 9.
• Step 2: Since the number ends in 0, it is essentially divisible by 10.
• Step 3: The prime factorization of 10 is $10 = 2 \times 5$.
• Step 4: Among these prime factors, we are asked which one surely appears. That is, any number ending in 0 will certainly include 5 as a prime factor.
• Step 5: Therefore, the prime factor that will surely appear is $5$.

Therefore, the solution to the problem is that the prime factor is $5$.

To solve this problem, we will make use of divisibility rules, particularly focusing on the rule for 2.

• Step 1: Analyze the given number form, $?12$, which indicates the number is a three-digit integer ending with the digits 12.
• Step 2: Apply divisibility rules. Note the last digit of $12$ is an even number, which is 2. By the rule of divisibility by 2, any number ending in an even number is divisible by 2.
• Step 3: Since the last digit is 2, it confirms the number is divisible by 2. Therefore, 2 is a prime factor of the number $?12$.

Given the options, $2$ is the only prime factor that will certainly appear among the first factors of any number ending with 12, as other numbers such as 3, 7, or 11 do not have guaranteed divisibility given non-fixed sum of digits or specific rules not directly applicable.

Therefore, the solution to the problem is $2$.

To solve this problem, let's identify the prime factor that is certain to be a part of a number in the format $3?0$. This number always ends with 0, indicating it is divisible by 10.

Step-by-step Solution:

• The number ends in 0: $3?0$. This means it is divisible by 10.
• Since a number divisible by 10 contains the prime factors of 10, which are 2 and 5, it means both 2 and 5 are factors of any such number.
• Out of the given prime factor options, 3, 5, 11, and 7, the prime factor 5 is certain to be present because every such number must end in 0, confirming divisibility by 5.

Thus, the prime factor that will surely appear among the first factors of a three-digit number in the format $3?0$ is $\mathbf{5}$.

We need to determine which prime factor is guaranteed for the number's appearance. The number is a three-digit number ending with 5, indicated as $??5$.

Based on divisibility rules, a number ending in 5 is divisible by 5. Therefore, 5 must be a factor.

Thus, the prime factor that surely appears in the factorization of any number ending in 5 is $5$.

Therefore, the solution to the problem is $5$.