The highest four-digit number which is divisible by each of the numbers 16, 36, 45, 48, is ?

Options

A. 9180
B. 9360
C. 9630
D. 9840

Answer

To find the highest four-digit number that is divisible by 16, 36, 45, and 48, we will follow these steps:

### Step 1: Find the LCM of the numbers
To determine the highest four-digit number that is divisible by each of the given numbers, we first need to find the Least Common Multiple (LCM) of 16, 36, 45, and 48.

**Prime Factorization:**
- 16 = 2^4
- 36 = 2^2 × 3^2
- 45 = 3^2 × 5^1
- 48 = 2^4 × 3^1

**Finding LCM:**
The LCM is obtained by taking the highest power of each prime factor:
- For 2: max(4, 2, 0, 4) = 4 → 2^4
- For 3: max(0, 2, 2, 1) = 2 → 3^2
- For 5: max(0, 0, 1, 0) = 1 → 5^1

Thus, the LCM is:
\[ \text{LCM} = 2^4 × 3^2 × 5^1 = 16 × 9 × 5 = 720 \]

### Step 2: Identify the highest four-digit number
The highest four-digit number is 9999.

### Step 3: Check divisibility by LCM
To find the highest four-digit number that is divisible by 720, we need to divide 9999 by 720 and find the largest integer quotient.

\[ \frac{9999}{720} \approx 13.88 \]

Taking the integer part, we have 13.

### Step 4: Calculate the highest four-digit number divisible by LCM
Now, we multiply the integer quotient by the LCM:

\[ 720 × 13 = 9360 \]

### Step 5: Verify divisibility
Finally, we check if 9360 is divisible by 16, 36, 45, and 48:
- 9360 ÷ 16 = 585 (divisible)
- 9360 ÷ 36 = 260 (divisible)
- 9360 ÷ 45 = 208 (divisible)
- 9360 ÷ 48 = 195 (divisible)

Since 9360 is divisible by all four numbers, we conclude that:

**The highest four-digit number which is divisible by 16, 36, 45, and 48 is 9360.**

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