What is the smallest number which leaves remainder 3 when divided by any of the numbers 5, 6 or 8 but leaves no remainder when it is divide by 9 ?

Options

A. 123
B. 603
C. 723
D. 243

Answer

To find the smallest number which leaves a remainder of 3 when divided by 5, 6, or 8, and leaves no remainder when divided by 9, we can follow these steps:

### Step 1: Understand the Remainder Condition
We need a number \( N \) such that:
- \( N \equiv 3 \mod 5 \)
- \( N \equiv 3 \mod 6 \)
- \( N \equiv 3 \mod 8 \)

This means that when \( N \) is divided by 5, 6, or 8, it leaves a remainder of 3.

### Step 2: Rewrite the Conditions
We can rewrite these conditions as:
- \( N - 3 \equiv 0 \mod 5 \)
- \( N - 3 \equiv 0 \mod 6 \)
- \( N - 3 \equiv 0 \mod 8 \)

Let \( M = N - 3 \). Then we need \( M \) to be a common multiple of 5, 6, and 8.

### Step 3: Find the Least Common Multiple (LCM)
Now we need to find the LCM of 5, 6, and 8:
- The prime factorization of 5 is \( 5^1 \).
- The prime factorization of 6 is \( 2^1 \times 3^1 \).
- The prime factorization of 8 is \( 2^3 \).

The LCM is calculated by taking the highest power of each prime:
- For \( 2 \): \( 2^3 \)
- For \( 3 \): \( 3^1 \)
- For \( 5 \): \( 5^1 \)

Thus, the LCM is:
\[ \text{LCM}(5, 6, 8) = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120 \]

### Step 4: Calculate \( N \)
Since \( M = N - 3 \), we have:
\[ M = 120k \quad \text{for some integer } k \]
Thus,
\[ N = 120k + 3 \]

### Step 5: Apply the Divisibility Condition
Now we need \( N \) to be divisible by 9:
\[ 120k + 3 \equiv 0 \mod 9 \]
Calculating \( 120 \mod 9 \):
\[ 120 \div 9 = 13 \quad \text{(remainder 3)} \]
So,
\[ 120 \equiv 3 \mod 9 \]
Thus,
\[ 120k + 3 \equiv 3k + 3 \equiv 0 \mod 9 \]
This simplifies to:
\[ 3(k + 1) \equiv 0 \mod 9 \]
This means \( k + 1 \) must be divisible by 3. Therefore, \( k + 1 = 3m \) for some integer \( m \), which gives us:
\[ k = 3m - 1 \]

### Step 6: Find the Smallest Positive \( N \)
To find the smallest positive \( N \), we can start with \( m = 1 \):
\[ k = 3(1) - 1 = 2 \]
Substituting back:
\[ N = 120(2) + 3 = 240 + 3 = 243 \]

### Conclusion
Thus, the smallest number which leaves a remainder of 3 when divided by 5, 6, or 8, and is divisible by 9 is:
\[ \boxed{243} \]