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

Options

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

Answer

To solve the problem, we need to find the smallest number that, when divided by 15, 20, or 35, leaves a remainder of 8. We can follow these steps:

### Step 1: Understand the Problem
We need to find a number \( N \) such that:
- \( N \mod 15 = 8 \)
- \( N \mod 20 = 8 \)
- \( N \mod 35 = 8 \)

This means that \( N - 8 \) must be divisible by 15, 20, and 35.

### Step 2: Set Up the Equation
Let \( M = N - 8 \). Then, we need to find \( M \) such that:
- \( M \mod 15 = 0 \)
- \( M \mod 20 = 0 \)
- \( M \mod 35 = 0 \)

This implies that \( M \) must be a common multiple of 15, 20, and 35.

### Step 3: Find the LCM of 15, 20, and 35
To find the least common multiple (LCM), we can use the prime factorization method:
- \( 15 = 3 \times 5 \)
- \( 20 = 2^2 \times 5 \)
- \( 35 = 5 \times 7 \)

Now, take the highest power of each prime:
- The highest power of 2 is \( 2^2 \) (from 20)
- The highest power of 3 is \( 3^1 \) (from 15)
- The highest power of 5 is \( 5^1 \) (from all)
- The highest power of 7 is \( 7^1 \) (from 35)

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

### Step 4: Calculate the LCM
Calculating step-by-step:
- \( 4 \times 3 = 12 \)
- \( 12 \times 5 = 60 \)
- \( 60 \times 7 = 420 \)

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

### Step 5: Find the Required Number
Now, since \( M = 420 \), we can find \( N \):
\[ N = M + 8 = 420 + 8 = 428 \]

### Conclusion
The smallest number \( N \) that, when divided by 15, 20, or 35, leaves a remainder of 8 is:
\[ \boxed{428} \]