A number when divided by 3, 4, 5 and 6 always leaves remainder of 2, but leaves no remainder when divided by 7. What is the lowest such number possible ?

Answer

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

1. When \( N \) is divided by 3, 4, 5, and 6, it leaves a remainder of 2.
2. When \( N \) is divided by 7, it leaves no remainder.

### Step-by-Step Solution:

**Step 1: Understand the conditions for the number \( N \)**
Since \( N \) leaves a remainder of 2 when divided by 3, 4, 5, and 6, we can express \( N \) in the following way:
\[ N = k \cdot \text{LCM}(3, 4, 5, 6) + 2 \]
where \( k \) is an integer.

**Step 2: Calculate the LCM of 3, 4, 5, and 6**
To find the least common multiple (LCM), we can break down each number into its prime factors:
- \( 3 = 3^1 \)
- \( 4 = 2^2 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \cdot 3^1 \)

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

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

**Step 3: Express \( N \)**
Now we can express \( N \) as:
\[ N = 60k + 2 \]

**Step 4: Apply the second condition (divisibility by 7)**
We need \( N \) to be divisible by 7:
\[ 60k + 2 \equiv 0 \mod 7 \]
This simplifies to:
\[ 60k \equiv -2 \mod 7 \]

**Step 5: Simplify \( 60 \mod 7 \)**
Calculating \( 60 \mod 7 \):
\[ 60 \div 7 = 8 \quad \text{(remainder 4)} \]
So,
\[ 60 \equiv 4 \mod 7 \]
Thus, we can rewrite the equation:
\[ 4k \equiv -2 \mod 7 \]
or equivalently,
\[ 4k \equiv 5 \mod 7 \] (since \(-2 \mod 7 = 5\))

**Step 6: Find the multiplicative inverse of 4 modulo 7**
To solve for \( k \), we need the multiplicative inverse of 4 modulo 7. We find \( x \) such that:
\[ 4x \equiv 1 \mod 7 \]
Testing values, we find:
- \( x = 2 \) works since \( 4 \cdot 2 = 8 \equiv 1 \mod 7 \).

**Step 7: Multiply both sides of the equation by the inverse**
Now, multiply both sides of \( 4k \equiv 5 \) by 2:
\[ k \equiv 10 \mod 7 \]
This simplifies to:
\[ k \equiv 3 \mod 7 \]

**Step 8: Find the smallest positive \( k \)**
The smallest non-negative integer \( k \) that satisfies this is \( k = 3 \).

**Step 9: Substitute \( k \) back to find \( N \)**
Now substitute \( k \) back into the equation for \( N \):
\[ N = 60 \cdot 3 + 2 = 180 + 2 = 182 \]

### Final Answer:
The lowest such number \( N \) is **182**.