Find the sum of digit of a smallest number which is divided by 5,6,7,8 leaves remainder 3 but when divided by 9 leaves no remainder .

Answer

To solve the problem, we need to find the smallest number \( N \) that satisfies the following conditions:

1. When divided by 5, 6, 7, and 8, it leaves a remainder of 3.
2. When divided by 9, it leaves no remainder.

### Step 1: Set up the equations based on the conditions

From the first condition, we can express \( N \) as:
\[ N = 5k + 3 \]
\[ N = 6m + 3 \]
\[ N = 7n + 3 \]
\[ N = 8p + 3 \]
for some integers \( k, m, n, p \).

This can be simplified to:
\[ N - 3 \equiv 0 \ (\text{mod} \ 5) \]
\[ N - 3 \equiv 0 \ (\text{mod} \ 6) \]
\[ N - 3 \equiv 0 \ (\text{mod} \ 7) \]
\[ N - 3 \equiv 0 \ (\text{mod} \ 8) \]

Let’s denote \( M = N - 3 \). Then we need \( M \) to be a common multiple of 5, 6, 7, and 8.

### Step 2: Find the least common multiple (LCM)

To find \( M \), we calculate the least common multiple (LCM) of 5, 6, 7, and 8.

- The prime factorization of the numbers:
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)
- \( 7 = 7^1 \)
- \( 8 = 2^3 \)

The LCM takes the highest power of each prime:
\[ \text{LCM}(5, 6, 7, 8) = 2^3 \times 3^1 \times 5^1 \times 7^1 = 8 \times 3 \times 5 \times 7 \]

Calculating this step-by-step:
\[ 8 \times 3 = 24 \]
\[ 24 \times 5 = 120 \]
\[ 120 \times 7 = 840 \]

Thus, \( \text{LCM}(5, 6, 7, 8) = 840 \).

### Step 3: Find \( N \)

Since \( M = 840 \), we have:
\[ N = M + 3 = 840 + 3 = 843 \]

### Step 4: Check the second condition

Now we check if \( N = 843 \) satisfies the second condition (divisible by 9):
\[ 843 \div 9 = 93 \quad \text{(exactly, with no remainder)} \]

### Step 5: Find the sum of the digits of \( N \)

Now we find the sum of the digits of \( N = 843 \):
\[ 8 + 4 + 3 = 15 \]

### Final Answer

The sum of the digits of the smallest number \( N \) is:
\[ \boxed{15} \]