Three brands A, B and C of biscuits are available in packets of 12, 15 and 21 biscuits respectively. If a shopkeeper wants to buy an equal number of biscuits, of each brand, what is the minimum number of packets of each brand, he should buy?

Answer

To solve the problem of how many packets of biscuits the shopkeeper should buy from brands A, B, and C, we need to find the least common multiple (LCM) of the number of biscuits in each packet. The packets contain 12, 15, and 21 biscuits respectively.

### Step-by-Step Solution:

1. **Identify the number of biscuits in each packet**:
- Brand A: 12 biscuits
- Brand B: 15 biscuits
- Brand C: 21 biscuits

2. **Find the prime factorization of each number**:
- 12 = 2 × 2 × 3 = \(2^2 \times 3^1\)
- 15 = 3 × 5 = \(3^1 \times 5^1\)
- 21 = 3 × 7 = \(3^1 \times 7^1\)

3. **Determine the highest power of each prime factor**:
- For 2: The highest power is \(2^2\) (from 12).
- For 3: The highest power is \(3^1\) (common in all).
- For 5: The highest power is \(5^1\) (from 15).
- For 7: The highest power is \(7^1\) (from 21).

4. **Calculate the LCM using the highest powers**:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 \]
\[ = 4 \times 3 \times 5 \times 7 \]
\[ = 12 \times 5 \times 7 \]
\[ = 60 \times 7 = 420 \]

5. **Determine the number of packets needed for each brand**:
- For Brand A (12 biscuits per packet):
\[ \text{Packets of A} = \frac{420}{12} = 35 \]
- For Brand B (15 biscuits per packet):
\[ \text{Packets of B} = \frac{420}{15} = 28 \]
- For Brand C (21 biscuits per packet):
\[ \text{Packets of C} = \frac{420}{21} = 20 \]

### Final Answer:
The shopkeeper should buy:
- 35 packets of Brand A
- 28 packets of Brand B
- 20 packets of Brand C