Find the least number which when divided by 2,3,4,5 and 6 leaves 1,2,3,4 and 5 as remainders respectively, but when divided by 7 leaves no remainders

Options

A. 210
B. 119
C. 126
D. 154

Answer

To solve the problem step by step, we need to find the least number that meets the specified conditions. Here’s how we can do it:

### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \mod 2 = 1 \)
- \( N \mod 3 = 2 \)
- \( N \mod 4 = 3 \)
- \( N \mod 5 = 4 \)
- \( N \mod 6 = 5 \)
- \( N \mod 7 = 0 \)

### Step 2: Set up the equations based on the remainders
From the conditions, we can rewrite them in terms of \( N \):
- \( N = 2k + 1 \) for some integer \( k \)
- \( N = 3m + 2 \) for some integer \( m \)
- \( N = 4n + 3 \) for some integer \( n \)
- \( N = 5p + 4 \) for some integer \( p \)
- \( N = 6q + 5 \) for some integer \( q \)

### Step 3: Find the least common multiple (LCM)
Notice that the remainders are always one less than the divisor. This means that \( N + 1 \) must be divisible by each of the divisors (2, 3, 4, 5, 6). Therefore, we need to find the LCM of these numbers:
- The LCM of 2, 3, 4, 5, and 6 is 60.

### Step 4: Express \( N \) in terms of \( k \)
Since \( N + 1 \) is a multiple of 60, we can express \( N \) as:
\[ N + 1 = 60k \]
Thus,
\[ N = 60k - 1 \]
where \( k \) is a natural number.

### Step 5: Apply the condition for divisibility by 7
Now, we need \( N \) to be divisible by 7:
\[ 60k - 1 \equiv 0 \mod 7 \]
This simplifies to:
\[ 60k \equiv 1 \mod 7 \]

### Step 6: Calculate \( 60 \mod 7 \)
Calculating \( 60 \mod 7 \):
\[ 60 \div 7 = 8 \quad \text{(remainder 4)} \]
So, \( 60 \equiv 4 \mod 7 \). Therefore, we need to solve:
\[ 4k \equiv 1 \mod 7 \]

### Step 7: Find the multiplicative inverse of 4 modulo 7
To find \( k \), we need the multiplicative inverse of 4 modulo 7. We can test values:
- \( 4 \times 1 = 4 \)
- \( 4 \times 2 = 8 \equiv 1 \mod 7 \)

Thus, \( k \equiv 2 \mod 7 \).

### Step 8: Find the smallest positive \( k \)
The smallest positive value of \( k \) that satisfies this condition is \( k = 2 \).

### Step 9: Substitute back to find \( N \)
Substituting \( k = 2 \) back into the equation for \( N \):
\[ N = 60(2) - 1 = 120 - 1 = 119 \]

### Conclusion
The least number which when divided by 2, 3, 4, 5, and 6 leaves remainders of 1, 2, 3, 4, and 5 respectively, and is divisible by 7 is:
\[ \boxed{119} \]