The least number which when divided by 2, 3, 4, 5 and 6 leaves the remainder 1 in each case. If the same number is divided by 7 it leaves no remainder. The number is :

Options

A. 231
B. 301
C. 371
D. 441

Answer

To find the least number which when divided by 2, 3, 4, 5, and 6 leaves a remainder of 1 in each case, and when divided by 7 leaves no remainder, we can follow these steps:

### Step 1: Determine the Least Common Multiple (LCM)
First, we need to find the least common multiple (LCM) of the numbers 2, 3, 4, 5, and 6.

- The prime factorization of each number is:
- 2 = 2
- 3 = 3
- 4 = 2^2
- 5 = 5
- 6 = 2 × 3

- To find the LCM, we take the highest power of each prime:
- Highest power of 2: 2^2 (from 4)
- Highest power of 3: 3^1 (from 3 or 6)
- Highest power of 5: 5^1 (from 5)

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

### Step 2: Adjust for Remainder
Since we need a number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, we can express our number as:
\[ N = 60k + 1 \]
where \( k \) is a non-negative integer.

### Step 3: Condition for Division by 7
Now, we also know that this number \( N \) must be divisible by 7. Therefore, we set up the equation:
\[ 60k + 1 \equiv 0 \ (\text{mod} \ 7) \]
This simplifies to:
\[ 60k \equiv -1 \ (\text{mod} \ 7) \]

### Step 4: Simplify the Modulo Condition
First, we find \( 60 \mod 7 \):
\[ 60 \div 7 = 8 \quad \text{(which gives a remainder of 4)} \]
Thus,
\[ 60 \equiv 4 \ (\text{mod} \ 7) \]

Now substituting into the equation:
\[ 4k \equiv -1 \ (\text{mod} \ 7) \]
This can be rewritten as:
\[ 4k \equiv 6 \ (\text{mod} \ 7) \]
(since \(-1\) is equivalent to \(6\) modulo \(7\)).

### Step 5: Find the Inverse of 4 Modulo 7
To solve for \( k \), we need the multiplicative inverse of 4 modulo 7. We can test values:
- \( 4 \times 2 = 8 \equiv 1 \ (\text{mod} \ 7) \)

Thus, the inverse of 4 is 2. Now we can multiply both sides of the equation by 2:
\[ k \equiv 2 \times 6 \ (\text{mod} \ 7) \]
\[ k \equiv 12 \equiv 5 \ (\text{mod} \ 7) \]

### Step 6: General Solution for k
The general solution for \( k \) is:
\[ k = 5 + 7m \]
where \( m \) is a non-negative integer.

### Step 7: Substitute Back to Find N
Now substituting \( k \) back into our equation for \( N \):
\[ N = 60(5 + 7m) + 1 \]
\[ N = 300 + 420m + 1 \]
\[ N = 301 + 420m \]

### Step 8: Find the Least Value of N
To find the least number, we set \( m = 0 \):
\[ N = 301 \]

### Conclusion
Thus, the least number which satisfies all conditions is:
\[ \boxed{301} \]