Prime and Composite Numbers Practice Problems Online

To determine whether 36 is a prime or composite number, we need to check if it has divisors other than 1 and 36:

• Step 1: Calculate the square root of 36, which is 6. This means we only need to test divisibility by numbers up to and including 6.
• Step 2: Check divisibility by 2. Since 36 is an even number (divisible by 2), it has a divisor other than 1 and itself.

Therefore, since 36 is divisible by 2 (and also by other numbers such as 3, 4, and more), it has divisors other than just 1 and 36. This means it cannot be a prime number.

Conclusively, the number 36 is Composite.

A number is classified as prime if it has exactly two distinct positive divisors: 1 and itself. Conversely, a number is composite if it has more than two divisors.

Given the number $n = 10$, we need to determine whether it is prime or composite.

Let's test the divisibility of 10 by numbers other than 1 and 10:

• Check divisibility by 2: Since 10 is an even number, it is divisible by 2. Specifically, $10 \div 2 = 5$ with no remainder.
• Check divisibility by 3: $10 \div 3 \approx 3.333$, which is not an integer, hence not divisible.
• Check divisibility by 5: $10 \div 5 = 2$ with no remainder, showing divisibility by 5.

Since 10 is divisible by numbers other than 1 and itself (specifically 2 and 5), it is not prime. Therefore, the number 10 is composite.

In conclusion, the number 10 is a composite number.

To determine if $n = 20$ is prime or composite, we need to examine its divisors.

• Step 1: Identify divisors of 20, other than 1 and 20 itself.
  Since 20 is an even number, it is divisible by 2. So, 2 is a divisor of 20.
• Step 2: Check divisibility by other small numbers:
  - 20 divided by 4 equals 5 (another divisor).
  - Additionally, $20 \div 5 = 4$, confirming 5 is also a divisor.

Since 20 has divisors other than 1 and itself (including 2, 4, and 5), it is not a prime number.

Therefore, the number $n = 20$ is Composite.

To determine whether $n = 23$ is a prime number, we will test its divisibility:

• Step 1: Calculate $\sqrt{23}$. The approximate value is 4.795, and thus we consider prime numbers up to the integer part, which is 4.
• Step 2: Check if 23 is divisible by any prime numbers less than or equal to 4. These primes are 2 and 3.

Step 3: Test divisibility:
- 23 is not divisible by 2, as it is odd.
- 23 is not divisible by 3, since $23 \div 3 \approx 7.67$, which is not an integer.

Since 23 is not divisible by any prime number less than or equal to its square root, it only has divisors of 1 and 23. Hence, 23 is a prime number.

Therefore, the solution to the problem is that $n = 23$ is prime.

To solve this problem, we'll determine if 42 is a prime or composite number by checking its divisibility by numbers other than 1 and itself.

A number is prime if it has exactly two distinct positive divisors: 1 and itself. It is composite if it has more than two distinct divisors.

Let's find the divisors of 42:

• $42 \div 1 = 42$
• $42 \div 2 = 21$ (evenly divisible, so 2 is a divisor)
• $42 \div 3 = 14$ (evenly divisible, so 3 is also a divisor)
• $42 \div 6 = 7$ (evenly divisible, so 6 is another divisor)
• $42 \div 7 = 6$ (evenly divisible, so 7 is a divisor)
• $42 \div 14 = 3$ (evenly divisible, so 14 is a divisor)
• $42 \div 21 = 2$ (evenly divisible, so 21 is a divisor)
• $42 \div 42 = 1$

From the above list, we can see that 42 has divisors other than 1 and itself, namely 2, 3, 6, 7, 14, and 21. This means that 42 is not a prime number.

Therefore, the number 42 is a composite number.