Find the smallest number which when divided by 2, 3, 4, 5 & 6 leaves remainder as 1 in each case & 0 when divided by 7?

Options

A. 121
B. 301
C. 403
D. 202

Answer

To find the smallest number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, and is divisible by 7, we can follow these steps:

### Step 1: Find the LCM of 2, 3, 4, 5, and 6
To find the least common multiple (LCM), we first determine the prime factorization of each number:

- 2 = 2
- 3 = 3
- 4 = 2^2
- 5 = 5
- 6 = 2 × 3

Now, we take the highest power of each prime number:

- The highest power of 2 is 2^2 (from 4)
- The highest power of 3 is 3^1 (from 3 or 6)
- The highest power of 5 is 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: Set up the equation
Since we need a number that leaves a remainder of 1 when divided by 2, 3, 4, 5, and 6, we can express this as:
\[ x \equiv 1 \mod 60 \]

This means that \( x \) can be written as:
\[ x = 60k + 1 \]
for some integer \( k \).

### Step 3: Find the condition for divisibility by 7
We also need \( x \) to be divisible by 7:
\[ 60k + 1 \equiv 0 \mod 7 \]

This simplifies to:
\[ 60k \equiv -1 \mod 7 \]

Calculating \( 60 \mod 7 \):
\[ 60 \div 7 = 8 \quad \text{(remainder 4)} \]
So,
\[ 60 \equiv 4 \mod 7 \]

Thus, we rewrite the equation:
\[ 4k \equiv -1 \mod 7 \]
or equivalently,
\[ 4k \equiv 6 \mod 7 \]
(since \(-1\) is equivalent to \(6\) in modulo \(7\)).

### Step 4: Solve for \( k \)
To solve \( 4k \equiv 6 \mod 7 \), we can find the multiplicative inverse of 4 modulo 7. The multiplicative inverse of 4 is 2 because:
\[ 4 \times 2 = 8 \equiv 1 \mod 7 \]

Now, multiply both sides of \( 4k \equiv 6 \) by 2:
\[ k \equiv 12 \mod 7 \]
which simplifies to:
\[ k \equiv 5 \mod 7 \]

This means:
\[ k = 7m + 5 \]
for some integer \( m \).

### Step 5: Substitute back to find \( x \)
Substituting \( k \) back into the equation for \( x \):
\[ x = 60(7m + 5) + 1 \]
\[ x = 420m + 300 + 1 \]
\[ x = 420m + 301 \]

### Step 6: Find the smallest \( x \)
To find the smallest positive \( x \), we set \( m = 0 \):
\[ x = 301 \]

### Conclusion
The smallest number which when divided by 2, 3, 4, 5, and 6 leaves a remainder of 1 in each case and is divisible by 7 is:
\[ \boxed{301} \]