Find the smallest number divisible by 15, 28 and 25 which gives remainder as 8 in each case.

Options

A. 2100
B. 2092
C. 2108
D. 2180

Answer

To find the smallest number divisible by 15, 28, and 25 that gives a remainder of 8 in each case, we can follow these steps:

### Step 1: Find the LCM of the numbers
We need to find the Least Common Multiple (LCM) of 15, 28, and 25.

- **Prime factorization**:
- 15 = 3 × 5
- 28 = 4 × 7 (or 2² × 7)
- 25 = 5²

- **Identify the highest power of each prime**:
- For 2: highest power is 2² (from 28)
- For 3: highest power is 3¹ (from 15)
- For 5: highest power is 5² (from 25)
- For 7: highest power is 7¹ (from 28)

- **Calculate the LCM**:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^2 \times 7^1 \]
\[ = 4 \times 3 \times 25 \times 7 \]
\[ = 4 \times 3 = 12 \]
\[ = 12 \times 25 = 300 \]
\[ = 300 \times 7 = 2100 \]

### Step 2: Adjust for the remainder
Since we need a number that gives a remainder of 8 when divided by 15, 28, and 25, we will add 8 to the LCM.

\[ \text{Smallest number} = \text{LCM} + 8 = 2100 + 8 = 2108 \]

### Conclusion
The smallest number that is divisible by 15, 28, and 25 and gives a remainder of 8 in each case is **2108**.

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