The least number, which when divided by 8, 9, 12 and 15 leaves 1 as a remainder in each case is?

Options

A. 179
B. 181
C. 359
D. 361

Answer

To find the least number which when divided by 8, 9, 12, and 15 leaves a remainder of 1, we can follow these steps:

### Step 1: Find the LCM of the numbers 8, 9, 12, and 15.
To find the LCM, we will first find the prime factorization of each number:

- **8** = \(2^3\)
- **9** = \(3^2\)
- **12** = \(2^2 \times 3^1\)
- **15** = \(3^1 \times 5^1\)

Now, we take the highest power of each prime factor:

- For \(2\), the highest power is \(2^3\) (from 8).
- For \(3\), the highest power is \(3^2\) (from 9).
- For \(5\), the highest power is \(5^1\) (from 15).

Thus, the LCM is calculated as follows:

\[ \text{LCM} = 2^3 \times 3^2 \times 5^1 \]

Calculating this gives:

\[ = 8 \times 9 \times 5 \]

Calculating step-by-step:

\[ 8 \times 9 = 72 \]
\[ 72 \times 5 = 360 \]

So, the LCM of 8, 9, 12, and 15 is **360**.

### Step 2: Since we need a number that leaves a remainder of 1 when divided by these numbers, we add 1 to the LCM.

\[ \text{Required number} = \text{LCM} + 1 = 360 + 1 = 361 \]

### Final Answer:
The least number which when divided by 8, 9, 12, and 15 leaves a remainder of 1 is **361**.

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