When a number is divided by 15 20 or 35, each time the remainder is 8. Then the samllest number is

Options

A. 428
B. 427
C. 328
D. 338

Answer

To solve the problem step by step, we need to find the smallest number that, when divided by 15, 20, or 35, leaves a remainder of 8.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find a number \( N \) such that:
- \( N \mod 15 = 8 \)
- \( N \mod 20 = 8 \)
- \( N \mod 35 = 8 \)

2. **Rewriting the Conditions**:
We can rewrite these conditions as:
- \( N - 8 \) is divisible by 15
- \( N - 8 \) is divisible by 20
- \( N - 8 \) is divisible by 35

This means that \( N - 8 \) is a common multiple of 15, 20, and 35.

3. **Finding the LCM**:
To find the smallest \( N - 8 \), we need to calculate the least common multiple (LCM) of 15, 20, and 35.

- The prime factorization of the numbers is:
- \( 15 = 3 \times 5 \)
- \( 20 = 2^2 \times 5 \)
- \( 35 = 5 \times 7 \)

- The LCM is found by taking the highest power of each prime:
- For \( 2 \): \( 2^2 \) (from 20)
- For \( 3 \): \( 3^1 \) (from 15)
- For \( 5 \): \( 5^1 \) (from all)
- For \( 7 \): \( 7^1 \) (from 35)

Therefore, the LCM is:
\[ \text{LCM} = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 \]

4. **Calculating the LCM**:
- First, calculate \( 4 \times 3 = 12 \)
- Then, \( 12 \times 5 = 60 \)
- Finally, \( 60 \times 7 = 420 \)

So, \( \text{LCM}(15, 20, 35) = 420 \).

5. **Finding the Smallest Number**:
Since \( N - 8 \) is the LCM, we have:
\[ N - 8 = 420 \]
Therefore, to find \( N \):
\[ N = 420 + 8 = 428 \]

### Final Answer:
The smallest number is **428**.