Let `x` be the largest 4-digit number which is divisible by each of 16, 21, 24 and 28. The sum of the digits of `x` is:

Options

A. 19
B. 21
C. 24
D. 16

Answer

To solve the problem step by step, we need to find the largest 4-digit number that is divisible by 16, 21, 24, and 28, and then calculate the sum of its digits.

### Step 1: Find the LCM of the numbers
To find the largest 4-digit number divisible by 16, 21, 24, and 28, we first need to calculate the least common multiple (LCM) of these numbers.

1. **Prime Factorization**:
- 16 = 2^4
- 21 = 3^1 × 7^1
- 24 = 2^3 × 3^1
- 28 = 2^2 × 7^1

2. **LCM Calculation**:
- Take the highest power of each prime factor:
- For 2: max(4, 3, 2) = 4 → 2^4
- For 3: max(1, 1) = 1 → 3^1
- For 7: max(1, 1) = 1 → 7^1
- Therefore, LCM = 2^4 × 3^1 × 7^1 = 16 × 3 × 7

3. **Calculate LCM**:
- 16 × 3 = 48
- 48 × 7 = 336
- Thus, LCM(16, 21, 24, 28) = 336.

### Step 2: Find the largest 4-digit number divisible by the LCM
The largest 4-digit number is 9999. To find the largest number less than or equal to 9999 that is divisible by 336, we perform the following steps:

1. **Divide 9999 by 336**:
- 9999 ÷ 336 ≈ 29.76 (we take the integer part which is 29).

2. **Multiply back to find the largest multiple**:
- 29 × 336 = 9732.

### Step 3: Calculate the sum of the digits of 9732
Now, we need to find the sum of the digits of 9732.

1. **Sum the digits**:
- 9 + 7 + 3 + 2 = 21.

### Final Answer
The sum of the digits of the largest 4-digit number divisible by 16, 21, 24, and 28 is **21**.