Find the greatest number of four digits which is exactly divisible by 48, 60 and 64.

Options

A. 9600
B. 9909
C. 9611
D. 9900

Answer

To find the greatest four-digit number that is exactly divisible by 48, 60, and 64, we will follow these steps:

### Step 1: Find the Least Common Multiple (LCM) of 48, 60, and 64
To determine the LCM, we first find the prime factorization of each number:

- **48**:
- 48 = 2^4 × 3^1
- **60**:
- 60 = 2^2 × 3^1 × 5^1
- **64**:
- 64 = 2^6

Next, we take the highest power of each prime factor from the factorizations:

- For **2**, the highest power is 2^6 (from 64).
- For **3**, the highest power is 3^1 (from both 48 and 60).
- For **5**, the highest power is 5^1 (from 60).

Now, we calculate the LCM:
\[ \text{LCM} = 2^6 \times 3^1 \times 5^1 \]
Calculating this:
\[ \text{LCM} = 64 \times 3 \times 5 = 1920 \]

### Step 2: Find the Greatest Four-Digit Number
The greatest four-digit number is 9999.

### Step 3: Check Divisibility
To find the greatest four-digit number that is divisible by 1920, we divide 9999 by 1920 and take the floor of the result:
\[ 9999 \div 1920 \approx 5.20625 \]
Taking the integer part, we have 5.

### Step 4: Multiply to Find the Largest Divisible Number
Now, we multiply 1920 by 5:
\[ 1920 \times 5 = 9600 \]

### Step 5: Verify Divisibility
Finally, we check if 9600 is divisible by 48, 60, and 64:
- **9600 ÷ 48 = 200** (exactly divisible)
- **9600 ÷ 60 = 160** (exactly divisible)
- **9600 ÷ 64 = 150** (exactly divisible)

Since 9600 is divisible by all three numbers, we conclude that the greatest four-digit number that is exactly divisible by 48, 60, and 64 is **9600**.

### Final Answer
The greatest four-digit number which is exactly divisible by 48, 60, and 64 is **9600**.

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