Prime Numbers and Composite Numbers: Complete the missing numbers

To solve this problem, we will follow these detailed steps:

• Step 1: Understand the definitions: A composite number has more than two divisors. A prime number has exactly two divisors.
• Step 2: Examine each choice to determine its type:
• $9$: Check divisibility by numbers other than 1 and itself. $9$ is divisible by 3, as $9 \div 3 = 3$. Thus, 9 is composite.
• $7$: Check divisibility. Only divides evenly by 1 and 7. Thus, 7 is prime.
• $3$: Check divisibility. Only divides evenly by 1 and 3. Thus, 3 is prime.
• $5$: Check divisibility. Only divides evenly by 1 and 5. Thus, 5 is prime.
• Step 3: Conclude which number is composite based on the divisor check:
• Since 9 is the only number with more than two divisors (1, 3, and 9), it is the composite number.

Therefore, the solution to the problem is $9$.

To solve this problem, we'll check each number to determine if it's a prime:

• Number 6: Greater than 1 but divisible by 2 and 3. Not prime.
• Number 4: Greater than 1 but divisible by 2. Not prime.
• Number 8: Greater than 1 but divisible by 2 and 4. Not prime.
• Number 2: Greater than 1 and not divisible by any other numbers except 1 and itself. It is prime.

Therefore, the prime number from the options is $2$.

To solve the problem, we will follow these steps:

• Step 1: List down the potential numbers we can form: 11, 51, 81, and 91.
• Step 2: Identify prime numbers by testing divisibility.

Let's analyze each number:

11: The only divisors of 11 are 1 and 11 itself, which makes it a prime number.

51: Check divisibility: 51 is divisible by 3, thus it is not prime because 51 ÷ 3 = 17.

81: Check divisibility: 81 is divisible by 3 (since 8+1=9, which is divisible by 3). So, 81 ÷ 3 = 27, and it is not a prime.

91: Check divisibility further: 91 is divisible by 7 (as 91 ÷ 7 = 13) which makes it not prime.

After examining each option, 11 is the only prime number.

Therefore, the solution to the problem is $11$.

To solve the problem of finding the missing digit in $\square2$ that results in a prime number, we need to check each possible digit from $0$ to $9$ and see which of them make $\square2$ a prime number.

Let's perform this step-by-step analysis:

• Step 1: Substitute $0$ in place of $\square$, resulting in the number $02$ which should be considered as $2$.
• Step 2: Check if the number $2$ is a prime number. A prime number is one that has no divisors other than $1$ and itself.
• Step 3: Determine if $2$ is prime. Since $2$ is divisible by only $1$ and $2$, it is a prime number.

Upon examining the possibilities, the use of $0$ in $\square$ results in $02$, which is equal to $2$, a prime number. Therefore, the missing digit that makes $\square2$ a prime number is $0$.

Thus, the correct number is $02$ or $2$, and therefore the correct choice from the given options is $0$.

To solve this problem, we'll proceed with the following steps:

• Step 1: Identify possible candidates for the digit filling $\square$.
• Step 2: Form numbers $29, 39, 59,$ and $79$ using the candidates $2, 3, 5,$ and $7$ respectively.
• Step 3: Test each number for composite status by checking for divisibility aside from 1 and itself.

Let's examine the numbers:

Step 1 and Step 2: Candidates give us the numbers $29, 39, 59,$ and $79$.

Step 3: Check each number:
- $29$ is only divisible by 1 and 29 (prime).
- $39$ is divisible by 1, 3, 13, and 39; hence, it is composite.
- $59$ is only divisible by 1 and 59 (prime).
- $79$ is only divisible by 1 and 79 (prime).

Therefore, the number $39$, formed by filling $\square$ with 3, is composite.

Thus, the correct number to fill in the blank is $3$.

To solve this problem, we'll fill in the missing digit and verify the primality of the constructed number:

1. List potential digits to fill in the square: These range from 0 to 9.
2. Apply the prime number test for each potential number:
   • $05$: Not a valid number, as it's less than 10.
   • $15$: Divisible by 3 ($15 \div 3 = 5$).
   • $25$: Divisible by 5.
   • $35$: Divisible by 5.
   • $45$: Divisible by 5.
   • $55$: Divisible by 5.
   • $65$: Divisible by 5.
   • $75$: Divisible by 5.
   • $85$: Divisible by 5.
   • $95$: Divisible by 5.
3. After testing all candidate numbers, only $\mathbf{05}$ was incorrectly formed as it is less than 10. All other numbers are non-prime because they have additional factors; numbers ending with 5 are divisible by 5.
4. Thus, only $\boxed{05}$, where the $\square$ is replaced by $0$, fits the requirement, resulting in a more sensible reading as simply 5, which is indeed prime.

Therefore, the solution to the problem is $0$.

To solve this problem, we'll conduct primality tests for each possible number formed by different digits in place of $\square$ in $\square7$.

• Step 1: Test if $37$ is prime.
• Step 2: Test if $57$ is prime.
• Step 3: Test if $87$ is prime.
• Step 4: Test if $77$ is prime.

Let's detail these steps:

Step 1: Check $37$.

$37$ is not divisible by any prime numbers up to its square root ($\sqrt{37} \approx 6.08$), specifically 2, 3, 5. Therefore, $37$ is prime.

Step 2: Check $57$.

$57$ is divisible by 3 ($57 \div 3 = 19$). Thus, $57$ is not prime.

Step 3: Check $87$.

$87$ is divisible by 3 ($87 \div 3 = 29$). Hence, $87$ is not prime.

Step 4: Check $77$.

$77$ is divisible by 7 ($77 \div 7 = 11$). Consequently, $77$ is not prime.

Therefore, the number that completes $\square7$ as a prime number is $\boxed{3}$, forming $37$ which is prime.