Find the smallest number which when divided by 35,56 and 91 leaves remainders 7.

Explanation

To find the smallest number which, when divided by 35, 56, and 91, leaves a remainder of 7, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \mod 35 = 7 \)
- \( N \mod 56 = 7 \)
- \( N \mod 91 = 7 \)

This means that \( N - 7 \) must be divisible by 35, 56, and 91.

### Step 2: Set up the equation
Let \( M = N - 7 \). Then, we need to find \( M \) such that:
- \( M \mod 35 = 0 \)
- \( M \mod 56 = 0 \)
- \( M \mod 91 = 0 \)

This means \( M \) is a common multiple of 35, 56, and 91. The smallest such \( M \) is the least common multiple (LCM) of these numbers.

### Step 3: Find the LCM of 35, 56, and 91
First, we find the prime factorization of each number:
- \( 35 = 5 \times 7 \)
- \( 56 = 2^3 \times 7 \)
- \( 91 = 7 \times 13 \)

Now, we take the highest power of each prime:
- The prime 2 appears in 56 as \( 2^3 \).
- The prime 5 appears in 35 as \( 5^1 \).
- The prime 7 appears in all three numbers, but the highest power is \( 7^1 \).
- The prime 13 appears in 91 as \( 13^1 \).

Thus, the LCM is:
\[ \text{LCM} = 2^3 \times 5^1 \times 7^1 \times 13^1 \]

### Step 4: Calculate the LCM
Calculating the LCM:
\[ \text{LCM} = 8 \times 5 \times 7 \times 13 \]

Calculating step-by-step:
1. \( 8 \times 5 = 40 \)
2. \( 40 \times 7 = 280 \)
3. \( 280 \times 13 = 3640 \)

So, \( \text{LCM}(35, 56, 91) = 3640 \).

### Step 5: Find \( N \)
Since \( M = N - 7 \), we have:
\[ N = M + 7 = 3640 + 7 = 3647 \]

### Final Answer
The smallest number which when divided by 35, 56, and 91 leaves a remainder of 7 is:
\[ \boxed{3647} \]