The least number which when divided by 4, 6, 8 and 9 leave zero remainder in each case and when divided by 13 leaves a remainder of 7 is:

Options

A. 144
B. 72
C. 36
D. 85

Answer

To find the least number which, when divided by 4, 6, 8, and 9 leaves a zero remainder, and when divided by 13 leaves a remainder of 7, we can follow these steps:

### Step 1: Find the LCM of 4, 6, 8, and 9
To find the least common multiple (LCM), we can use the prime factorization method.

- **Prime factorization:**
- 4 = \(2^2\)
- 6 = \(2^1 \times 3^1\)
- 8 = \(2^3\)
- 9 = \(3^2\)

- **Take the highest power of each prime:**
- For \(2\), the highest power is \(2^3\) (from 8).
- For \(3\), the highest power is \(3^2\) (from 9).

- **Calculate the LCM:**
\[ \text{LCM} = 2^3 \times 3^2 = 8 \times 9 = 72 \]

### Step 2: Express the number in terms of LCM
Let the number be \(N\). Since \(N\) must be a multiple of the LCM, we can express it as:
\[ N = 72k \quad \text{(where \(k\) is a positive integer)} \]

### Step 3: Set up the equation for the remainder condition
According to the problem, when \(N\) is divided by 13, it should leave a remainder of 7. Therefore, we can write:
\[ 72k \equiv 7 \mod 13 \]

### Step 4: Calculate \(72 \mod 13\)
First, we need to find \(72 \mod 13\):
\[ 72 \div 13 = 5 \quad \text{(which gives a quotient of 5)} \]
\[ 72 - (13 \times 5) = 72 - 65 = 7 \]
Thus,
\[ 72 \equiv 7 \mod 13 \]

### Step 5: Substitute back into the equation
Now substituting back, we have:
\[ 7k \equiv 7 \mod 13 \]

### Step 6: Solve for \(k\)
To solve for \(k\), we can divide both sides by 7. Since 7 has a multiplicative inverse modulo 13, we can simplify:
\[ k \equiv 1 \mod 13 \]
This means \(k\) can be expressed as:
\[ k = 1 + 13m \quad \text{(where \(m\) is a non-negative integer)} \]

### Step 7: Find the least value of \(N\)
Substituting \(k = 1\) (the smallest value for \(k\)):
\[ N = 72 \times 1 = 72 \]

### Step 8: Check the condition for \(N\)
Now we check if \(N = 72\) satisfies the condition when divided by 13:
\[ 72 \div 13 = 5 \quad \text{(remainder 7)} \]
This confirms that \(72\) leaves a remainder of 7 when divided by 13.

### Conclusion
Thus, the least number which meets the criteria is:
\[ \boxed{72} \]