The least number which when decreased by 9 is exactly divisible by 12, 16, 24 and 48 is ______

Options

A. 16
B. 48
C. 57
D. 39

Answer

To find the least number which when decreased by 9 is exactly divisible by 12, 16, 24, and 48, we can follow these steps:

### Step 1: Find the Least Common Multiple (LCM)
First, we need to find the least common multiple (LCM) of the numbers 12, 16, 24, and 48.

- The prime factorization of each number is:
- 12 = 2^2 * 3^1
- 16 = 2^4
- 24 = 2^3 * 3^1
- 48 = 2^4 * 3^1

To find the LCM, we take the highest power of each prime factor:
- For 2, the highest power is 2^4 (from 16 or 48).
- For 3, the highest power is 3^1 (from 12, 24, or 48).

Thus, the LCM is:
\[ \text{LCM} = 2^4 * 3^1 = 16 * 3 = 48 \]

### Step 2: Set Up the Equation
Let the least number be \( x \). According to the problem, we have:
\[ x - 9 \text{ is divisible by } 48 \]
This can be expressed as:
\[ x - 9 = 48k \]
for some integer \( k \).

### Step 3: Solve for \( x \)
Rearranging the equation gives us:
\[ x = 48k + 9 \]

### Step 4: Find the Least Value of \( x \)
To find the least number \( x \), we can start with the smallest integer value for \( k \), which is 1:
\[ x = 48(1) + 9 = 48 + 9 = 57 \]

### Conclusion
Thus, the least number which when decreased by 9 is exactly divisible by 12, 16, 24, and 48 is:
\[ \boxed{57} \]

---