Find the smallest number which is divided by 2,3,4,5,6 and leaves remainder 1 in each case but number is exactly divisible by 7.

Answer

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

1. When \( N \) is divided by 2, 3, 4, 5, and 6, it leaves a remainder of 1.
2. \( N \) is exactly divisible by 7.

### Step 1: Understand the first condition
If \( N \) leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, we can express this mathematically as:
\[ N \equiv 1 \ (\text{mod} \ 2) \]
\[ N \equiv 1 \ (\text{mod} \ 3) \]
\[ N \equiv 1 \ (\text{mod} \ 4) \]
\[ N \equiv 1 \ (\text{mod} \ 5) \]
\[ N \equiv 1 \ (\text{mod} \ 6) \]

This means that \( N - 1 \) must be divisible by each of these numbers. Therefore, we can find the least common multiple (LCM) of these numbers.

### Step 2: Calculate the LCM of 2, 3, 4, 5, and 6
- The prime factorization of each number is:
- \( 2 = 2^1 \)
- \( 3 = 3^1 \)
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \times 3^1 \)

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

Thus, the LCM is:
\[ \text{LCM}(2, 3, 4, 5, 6) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60 \]

### Step 3: Express \( N \)
Since \( N - 1 \) must be a multiple of 60, we can express \( N \) as:
\[ N = 60k + 1 \]
for some integer \( k \).

### Step 4: Apply the second condition
We also need \( N \) to be divisible by 7:
\[ 60k + 1 \equiv 0 \ (\text{mod} \ 7) \]
This simplifies to:
\[ 60k \equiv -1 \ (\text{mod} \ 7) \]
Calculating \( 60 \mod 7 \):
\[ 60 \div 7 = 8 \quad \text{(remainder 4)} \]
Thus:
\[ 60 \equiv 4 \ (\text{mod} \ 7) \]
So we have:
\[ 4k \equiv -1 \ (\text{mod} \ 7) \]
or equivalently:
\[ 4k \equiv 6 \ (\text{mod} \ 7) \]

### Step 5: Solve for \( k \)
To solve \( 4k \equiv 6 \ (\text{mod} \ 7) \), we can find the multiplicative inverse of 4 modulo 7. Testing values, we find:
- \( 4 \times 2 = 8 \equiv 1 \ (\text{mod} \ 7) \)

Thus, the inverse is 2. Multiplying both sides of the equation by 2:
\[ k \equiv 12 \ (\text{mod} \ 7) \]
\[ k \equiv 5 \ (\text{mod} \ 7) \]

### Step 6: Find the smallest \( k \)
The smallest non-negative integer satisfying this is \( k = 5 \).

### Step 7: Calculate \( N \)
Substituting \( k \) back into the equation for \( N \):
\[ N = 60 \times 5 + 1 = 300 + 1 = 301 \]

### Conclusion
The smallest number \( N \) that satisfies both conditions is:
\[ \boxed{301} \]