Question:

Given: Two numbers 96 and 404.

Step 1 – Prime factorisation
\[96 = 2 \times 48 = 2 \times 2 \times 24 = 2 \times 2 \times 2 \times 12 = 2 \times 2 \times 2 \times 2 \times 6 = 2^5 \times 3\]
\[404 = 2 \times 202 = 2 \times 2 \times 101 = 2^2 \times 101\]
(Here 101 is a prime number.)

Step 2 – Find HCF
Identify the common prime factors and take the smallest exponent of each:
- Common prime factor: \(2\)
- Smallest exponent of \(2\) in the two factorizations = \(2\) (since \(96 = 2^5\) and \(404 = 2^2\)).
Thus,
\[\text{HCF} = 2^{2} = 4\]

Step 3 – Find LCM
Take each prime factor appearing in either number with the highest exponent:
- \(2\) : highest exponent = \(5\) (from 96)
- \(3\) : appears only in 96 → exponent = \(1\)
- \(101\) : appears only in 404 → exponent = \(1\)
Hence,
\[\text{LCM} = 2^{5} \times 3^{1} \times 101^{1} = 32 \times 3 \times 101 = 96 \times 101 = 9696\]

Step 4 – Verification (optional)
Using the relation \(\text{LCM} \times \text{HCF} = a \times b\):
\[9696 \times 4 = 96 \times 404 = 38784\]
Both sides are equal, confirming the results.

Conclusion
- HCF of 96 and 404 = 4
- LCM of 96 and 404 = 9696