Find the greatest number of 3 digits which when divided by 6,9,12 leaves 3 as remainder in each case.

Options

A. 975
B. 996
C. 903
D. 936

Answer

To find the greatest three-digit number that leaves a remainder of 3 when divided by 6, 9, and 12, we can follow these steps:

### Step 1: Understand the Problem
We need to find a three-digit number \( N \) such that:
- \( N \mod 6 = 3 \)
- \( N \mod 9 = 3 \)
- \( N \mod 12 = 3 \)

This means that \( N - 3 \) must be divisible by 6, 9, and 12.

### Step 2: Calculate the Least Common Multiple (LCM)
To find a number that is divisible by 6, 9, and 12, we need to calculate the LCM of these numbers.

- The prime factorization of the numbers is:
- \( 6 = 2^1 \times 3^1 \)
- \( 9 = 3^2 \)
- \( 12 = 2^2 \times 3^1 \)

- The LCM is found by taking the highest power of each prime:
- For \( 2 \): the highest power is \( 2^2 \) (from 12)
- For \( 3 \): the highest power is \( 3^2 \) (from 9)

Thus, the LCM is:
\[ \text{LCM}(6, 9, 12) = 2^2 \times 3^2 = 4 \times 9 = 36 \]

### Step 3: Find the Greatest Three-Digit Number
The greatest three-digit number is 999. We need to find the largest number less than or equal to 999 that, when reduced by 3, is divisible by 36.

### Step 4: Calculate the Largest Multiple of 36 Less Than or Equal to 996
Since \( N - 3 \) must be divisible by 36, we can set:
\[ N - 3 = 36k \quad \text{for some integer } k \]
Thus, we can rewrite \( N \) as:
\[ N = 36k + 3 \]

To find the largest \( k \) such that \( N \) is a three-digit number:
\[ 36k + 3 \leq 999 \]
Subtracting 3 from both sides:
\[ 36k \leq 996 \]
Dividing both sides by 36:
\[ k \leq \frac{996}{36} \approx 27.6667 \]
Thus, the largest integer \( k \) is 27.

### Step 5: Calculate \( N \)
Now substituting \( k = 27 \):
\[ N = 36 \times 27 + 3 = 972 + 3 = 975 \]

### Conclusion
The greatest three-digit number that leaves a remainder of 3 when divided by 6, 9, and 12 is **975**.

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