Three brands B1, B2 and B3 of shirts are available in cartons of 12, 15 and 21 shirts respectively. A stockist wants to stock the same number of shirts of each brand in his warehouse. What is the minimum number of cartons of each brand that he nneds to buy?

Answer

To solve the problem of how many cartons of each brand of shirts the stockist needs to buy, we will follow these steps:

### Step 1: Identify the number of shirts in each carton
- Brand B1 has 12 shirts per carton.
- Brand B2 has 15 shirts per carton.
- Brand B3 has 21 shirts per carton.

### Step 2: Find the Least Common Multiple (LCM) of the carton sizes
To find the minimum number of shirts that can be equally stocked from each brand, we need to calculate the LCM of 12, 15, and 21.

#### Prime Factorization:
- **12**: \( 2^2 \times 3^1 \)
- **15**: \( 3^1 \times 5^1 \)
- **21**: \( 3^1 \times 7^1 \)

#### LCM Calculation:
The LCM is found by taking the highest power of each prime factor:
- For \(2\): \(2^2\) (from 12)
- For \(3\): \(3^1\) (common in all)
- For \(5\): \(5^1\) (from 15)
- For \(7\): \(7^1\) (from 21)

Now, we multiply these together:
\[ LCM = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 \]

Calculating step by step:
1. \(4 \times 3 = 12\)
2. \(12 \times 5 = 60\)
3. \(60 \times 7 = 420\)

Thus, the LCM of 12, 15, and 21 is **420**.

### Step 3: Calculate the number of cartons needed for each brand
Now we will divide the LCM by the number of shirts in each carton to find out how many cartons are needed for each brand.

- For **Brand B1** (12 shirts per carton):
\[ \text{Number of cartons} = \frac{420}{12} = 35 \]

- For **Brand B2** (15 shirts per carton):
\[ \text{Number of cartons} = \frac{420}{15} = 28 \]

- For **Brand B3** (21 shirts per carton):
\[ \text{Number of cartons} = \frac{420}{21} = 20 \]

### Final Answer:
- Brand B1: 35 cartons
- Brand B2: 28 cartons
- Brand B3: 20 cartons