What is the least number which when divided by the number 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 22 leaves no remainder?

Options

A. 312
B. 242
C. 1562
D. 1586

Answer

To solve the problem step by step, we need to find the least number that meets the specified conditions.

### Step 1: Understand the conditions
We need to find a number \( N \) such that:
1. When \( N \) is divided by 3, 5, 6, 8, 10, and 12, it leaves a remainder of 2.
2. When \( N \) is divided by 22, it leaves no remainder.

### Step 2: Set up the equation for the first condition
Since \( N \) leaves a remainder of 2 when divided by each of these numbers, we can express \( N \) in the form:
\[ N = k \cdot \text{LCM}(3, 5, 6, 8, 10, 12) + 2 \]
where \( k \) is some integer.

### Step 3: Calculate the LCM
To find the least common multiple (LCM) of the numbers 3, 5, 6, 8, 10, and 12, we can break each number down into its prime factors:
- \( 3 = 3^1 \)
- \( 5 = 5^1 \)
- \( 6 = 2^1 \cdot 3^1 \)
- \( 8 = 2^3 \)
- \( 10 = 2^1 \cdot 5^1 \)
- \( 12 = 2^2 \cdot 3^1 \)

Now, we take the highest power of each prime:
- For \( 2 \): the highest power is \( 2^3 \) (from 8).
- For \( 3 \): the highest power is \( 3^1 \) (from 3, 6, and 12).
- For \( 5 \): the highest power is \( 5^1 \) (from 5 and 10).

Thus, the LCM is:
\[ \text{LCM} = 2^3 \cdot 3^1 \cdot 5^1 = 8 \cdot 3 \cdot 5 = 120. \]

### Step 4: Substitute LCM into the equation
Now we can substitute the LCM back into our equation for \( N \):
\[ N = 120k + 2. \]

### Step 5: Apply the second condition
Now we need \( N \) to be divisible by 22:
\[ 120k + 2 \equiv 0 \mod 22. \]

### Step 6: Simplify the equation
We can simplify this to find \( k \):
\[ 120k + 2 \equiv 0 \mod 22 \]
This can be rewritten as:
\[ 120k \equiv -2 \mod 22. \]
Calculating \( 120 \mod 22 \):
\[ 120 \div 22 = 5 \quad \text{(remainder 10)} \]
So,
\[ 120 \equiv 10 \mod 22. \]
Now substituting back:
\[ 10k \equiv -2 \mod 22. \]
This simplifies to:
\[ 10k \equiv 20 \mod 22. \]

### Step 7: Solve for \( k \)
To solve for \( k \), we can divide both sides by 2 (since 10 and 20 are both even):
\[ 5k \equiv 10 \mod 11. \]
Now, dividing by 5:
\[ k \equiv 2 \mod 11. \]

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

### Step 9: Calculate \( N \)
Now substitute \( k = 2 \) back into the equation for \( N \):
\[ N = 120 \cdot 2 + 2 = 240 + 2 = 242. \]

### Conclusion
Thus, the least number which meets the conditions is:
\[ \boxed{242}. \]