Prime Numbers & Factorization Sums for Class 6 — Free CBSE Worksheet PDF with Answers

Think of prime numbers as the atoms of mathematics. Just like everything in the universe is built from atoms, every whole number greater than $1$ is either a prime number or can be broken down into a product of primes. So $12 = 2 \times 2 \times 3$ and $45 = 3 \times 3 \times 5$. This breakdown is called prime factorisation, and it is the central idea of the Class 6 CBSE Prime Time chapter (Ganita Prakash Chapter 5).

A prime number has exactly two factors: $1$ and itself. The smallest prime is $2$, and it is also the only even prime. A composite number has more than two factors. The number $1$ is neither prime nor composite — it has only one factor, itself.

This worksheet has 60 questions across three levels. Level 1 covers identifying primes and composites, listing factors and multiples, and divisibility checks. Level 2 builds prime factorisation skills and HCF/LCM by listing. Level 3 handles HCF and LCM by prime factorisation, plus word problems involving bells, intervals, and arrangements.

There are exactly 25 prime numbers below 100: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97$. The Sieve of Eratosthenes is the cleanest way to find them. Write down numbers $1$ to $100$. Cross out $1$ (not prime). Circle $2$, then cross out every multiple of $2$. Circle the next uncrossed number ($3$), then cross out every multiple of $3$. Repeat with $5$ and $7$. Whatever is left circled is prime.

Why does it stop at 7? Because $\sqrt{100} = 10$, and any composite number under 100 must have a factor under 10. Once you have crossed out multiples of all primes below 10 (which are $2, 3, 5, 7$), the remaining circled numbers are guaranteed prime. This is a CBSE Class 6 staple exercise.

To find the prime factorisation of $120$: split into any two factors, say $120 = 8 \times 15$. Then split each: $8 = 2 \times 4$ and $15 = 3 \times 5$. Continue: $4 = 2 \times 2$. Now all branches end in primes. Read off the leaves: $120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$.

Whether you start by splitting $120$ as $2 \times 60$ or $10 \times 12$ or $8 \times 15$ does not matter. The prime factorisation is unique — every path leads to the same answer $2^3 \times 3 \times 5$. This is called the Fundamental Theorem of Arithmetic, and it is one of the most important ideas in number theory.

To find HCF (Highest Common Factor), write the prime factorisation of each number, then for every prime, take the smallest power that appears. For $24 = 2^3 \times 3$ and $36 = 2^2 \times 3^2$: HCF $= 2^2 \times 3 = 12$. (Smallest power of 2 is $2^2$; smallest power of 3 is $3^1$.)

To find LCM (Lowest Common Multiple), take the largest power of each prime that appears. For the same numbers: LCM $= 2^3 \times 3^2 = 72$. There is also a useful identity: HCF $\times$ LCM $=$ product of the two numbers. Check: $12 \times 72 = 864 = 24 \times 36$. Correct.

Word problem example: two bells ring at intervals of 15 minutes and 20 minutes. If both ring at 9:00 AM, when do they next ring together? Find LCM of 15 and 20: $15 = 3 \times 5$, $20 = 2^2 \times 5$, so LCM $= 2^2 \times 3 \times 5 = 60$. They both ring together every 60 minutes — next at 10:00 AM.