JEE Mathematics Prime and Composite Numbers Notes Set 01

Prime and Composite Numbers

A prime number is a natural number greater than 1 whose only divisors are 1 and itself: 2, 3, 5, 7, 11, 13, 17, … The number 2 is the only even prime. A natural number greater than 1 that is not prime is composite, for example 4, 6, 9, 15. The number 1 is neither prime nor composite.

Unique Prime Factorisation

Every composite number can be written as a product of primes, and the factorisation is unique apart from the order of the factors. For example 150 = 2 × 3 × 5². Dividing repeatedly by primes (150 ÷ 2 = 75, 75 ÷ 3 = 25, 25 ÷ 5 = 5) finds it. If N = p^a q^b r^c with p, q, r distinct primes, then N has (a + 1)(b + 1)(c + 1) positive divisors. For 150 this is 2 × 2 × 3 = 12.

Products of Consecutive Integers

1. The product of two consecutive integers is divisible by 2 (for example 36 × 37).
2. The product of three consecutive integers is divisible by 6 (for example 21 × 22 × 23 contains the factors 3 and 2 × 11).
3. In general the product of r consecutive integers is divisible by r!.

Proving a Number is Irrational

Irrationality of \( \sqrt{2} \): suppose \( \sqrt{2} = \frac{p}{q} \) in lowest terms. Then p² = 2q², so p² is even, hence p is even, say p = 2m. Then 4m² = 2q², so q² = 2m² and q is even too. Both p and q are even, contradicting lowest terms, so \( \sqrt{2} \) is irrational. Once this is known, 1 + \( \sqrt{2} \) is irrational because \( \sqrt{2} = (1 + \sqrt{2}) - 1 \) would otherwise be rational.

Worked Examples

Example 1. It is given that p, p + 2 and p + 4 are all prime. Find p.
Modulo 3 the numbers p, p + 2, p + 4 leave the three different remainders 0, 1, 2 in some order, so exactly one of them is divisible by 3. A prime divisible by 3 must be 3 itself. If p + 2 = 3 or p + 4 = 3 then p would be 1 or −1, which is not prime, so p = 3 and the primes are 3, 5, 7.

Example 2. Find the number of positive integral values of x satisfying ax = a + 10x for integer a.
Rearranging, x(a − 10) = a, so \( x = \frac{a}{a - 10} = 1 + \frac{10}{a - 10} \). For x to be a positive integer, a − 10 must be a positive divisor of 10 (a negative divisor gives x ≤ 0). The divisors 1, 2, 5, 10 give x = 11, 6, 3, 2, so there are 4 values.

Example 3. A is the set of even numbers between 5 and 13, B the set of divisors of 70 and C the set of primes below 17. Then A = {6, 8, 10, 12}, B = {1, 2, 5, 7, 10, 14, 35, 70}, C = {2, 3, 5, 7, 11, 13}, and B ∩ C = {2, 5, 7}, while A ∩ (B ∩ C) = ∅.