What is the least number, divided by 12, 16, 24 and 36 gives remainder 9 in each case?

Options

A. 151
B. 149
C. 153
D. 137

Answer

To find the least number that, when divided by 12, 16, 24, and 36, gives a remainder of 9 in each case, we can follow these steps:

### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \mod 12 = 9 \)
- \( N \mod 16 = 9 \)
- \( N \mod 24 = 9 \)
- \( N \mod 36 = 9 \)

### Step 2: Rewrite the equations
From the equations above, we can rewrite them as:
- \( N = 12k + 9 \) for some integer \( k \)
- \( N = 16m + 9 \) for some integer \( m \)
- \( N = 24n + 9 \) for some integer \( n \)
- \( N = 36p + 9 \) for some integer \( p \)

This means that \( N - 9 \) must be divisible by 12, 16, 24, and 36.

### Step 3: Find the least common multiple (LCM)
To find the smallest \( N - 9 \), we need to calculate the LCM of the divisors:
- The prime factorization of each number:
- \( 12 = 2^2 \times 3^1 \)
- \( 16 = 2^4 \)
- \( 24 = 2^3 \times 3^1 \)
- \( 36 = 2^2 \times 3^2 \)

The LCM will take the highest power of each prime:
- For \( 2 \): highest power is \( 2^4 \) (from 16)
- For \( 3 \): highest power is \( 3^2 \) (from 36)

Thus, the LCM is:
\[ LCM = 2^4 \times 3^2 = 16 \times 9 = 144 \]

### Step 4: Calculate \( N \)
Now, since \( N - 9 \) must be a multiple of 144, we can express \( N \) as:
\[ N = 144k + 9 \]
where \( k \) is a non-negative integer. To find the least number, we set \( k = 1 \):
\[ N = 144 \times 1 + 9 = 153 \]

### Step 5: Conclusion
The least number that, when divided by 12, 16, 24, and 36, gives a remainder of 9 is:
\[ \boxed{153} \]