Important Problems on Prime Numbers

This section on Prime Numbers features a collection of important and frequently asked problems designed to help students master the topic through practice. Covering a range of difficulty levels, these problems focus on identifying prime numbers, applying divisibility rules, understanding prime factorization, and exploring patterns in prime distribution. Each problem is accompanied by a clear, step-by-step solution to support learning and concept reinforcement. Ideal for students preparing for exams or math competitions, this guide enhances problem-solving skills and deepens your understanding of Prime Numbers.

If an integer \( a > 1 \) is not divisible by any prime \( p \leq \sqrt{a} \), then prove that \( a \) is a prime.

Answer:

We will prove the contrapositive: If \( a \) is not prime (i.e., composite), then it is divisible by some prime \( p \leq \sqrt{a} \).

Proof:

Assume that \( a > 1 \) is a composite number. Then, by definition, \( a = m \cdot n \) for some integers \( m, n \) such that \( 1 < m \leq n < a \).

Now consider:

a = m \cdot n \Rightarrow m \leq \sqrt{a} \quad \text{or} \quad n \leq \sqrt{a}

This inequality holds because if both \( m > \sqrt{a} \) and \( n > \sqrt{a} \), then:

a = m \cdot n > \sqrt{a} \cdot \sqrt{a} = a

which is a contradiction. So at least one of the factors \( m \) or \( n \) is \( \leq \sqrt{a} \).

Without loss of generality, let \( m \leq \sqrt{a} \). Now, since \( m > 1 \), it must have a prime divisor (by the Fundamental Theorem of Arithmetic). Let this prime be \( p \), so \( p \mid m \), and hence \( p \mid a \).

Also, since \( p \mid m \) and \( m \leq \sqrt{a} \), it follows that:

p \leq m \leq \sqrt{a}

This means \( a \) is divisible by some prime \( p \leq \sqrt{a} \).

Hence, if \( a \) is not divisible by any prime \( \leq \sqrt{a} \), it cannot be composite and must be prime.

Find all prime numbers that divide \(50!\).

Answer:

To find all prime numbers that divide \(50!\), we recall that:

50! = 1 \cdot 2 \cdot 3 \cdot 4 \cdot \cdots \cdot 50

So \(50!\) is the product of all positive integers from 1 to 50. Any prime number \(p \leq 50\) appears as a factor in this product, and thus divides \(50!\).

Therefore, all prime numbers less than or equal to 50 divide \(50!\).

List of all primes \( \leq 50 \):

2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\ 23,\ 29,\ 31,\ 37,\ 41,\ 43,\ 47

Conclusion:
The prime numbers that divide \(50!\) are:

2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\ 23,\ 29,\ 31,\ 37,\ 41,\ 43,\ 47