The least number of 5-digits which is divisible by 45, 60, 75 and 120 when it is added to 119?

Options

A. 10800
B. 10681
C. 10321
D. 10941

Answer

To solve the problem of finding the least 5-digit number that is divisible by 45, 60, 75, and 120, and then adding 119 to it, we can follow these steps:

### Step 1: Find the LCM of the numbers
To find the least number that is divisible by 45, 60, 75, and 120, we first need to calculate the Least Common Multiple (LCM) of these numbers.

- **Prime Factorization**:
- 45 = 3^2 × 5^1
- 60 = 2^2 × 3^1 × 5^1
- 75 = 3^1 × 5^2
- 120 = 2^3 × 3^1 × 5^1

- **Take the highest power of each prime**:
- For 2: max(0, 2, 0, 3) = 3 → 2^3
- For 3: max(2, 1, 1, 1) = 2 → 3^2
- For 5: max(1, 1, 2, 1) = 2 → 5^2

So, the LCM = 2^3 × 3^2 × 5^2 = 8 × 9 × 25 = 1800.

### Step 2: Find the smallest 5-digit number
The smallest 5-digit number is 10000.

### Step 3: Find the smallest 5-digit number divisible by LCM
To find the smallest 5-digit number that is divisible by 1800, we divide 10000 by 1800 and round up to the nearest whole number:

- 10000 ÷ 1800 ≈ 5.555... → Round up to 6.

Now, multiply by 1800 to find the smallest 5-digit number:

- 6 × 1800 = 10800.

### Step 4: Add 119 to the result
Now we add 119 to the smallest 5-digit number we found:

- 10800 + 119 = 10919.

### Final Answer
The least number of 5-digits which is divisible by 45, 60, 75, and 120 when added to 119 is **10919**.

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