What is the greatest number of 4-digits that which when divided by any of the numbers 6,9,12 and 17 leaves a remainder of 1?

Options

A. 9997
B. 9793
C. 9895
D. 9487

Answer

To find the greatest 4-digit number that leaves a remainder of 1 when divided by 6, 9, 12, and 17, we can follow these steps:

### Step 1: Find the LCM of the given numbers
We need to find the least common multiple (LCM) of 6, 9, 12, and 17.

- The prime factorization of each number is:
- 6 = 2 × 3
- 9 = 3²
- 12 = 2² × 3
- 17 = 17 (17 is a prime number)

- The LCM is found by taking the highest power of each prime:
- For 2: the highest power is 2² (from 12)
- For 3: the highest power is 3² (from 9)
- For 17: the highest power is 17¹ (from 17)

Thus, the LCM = 2² × 3² × 17 = 4 × 9 × 17 = 612.

### Step 2: Identify the largest 4-digit number
The largest 4-digit number is 9999.

### Step 3: Find the largest multiple of the LCM less than or equal to 9999
We need to divide 9999 by the LCM (612) to find the largest multiple of 612 that is less than or equal to 9999.

- Performing the division:
\[ 9999 ÷ 612 \approx 16.33 \]
- The largest integer less than or equal to 16.33 is 16.

### Step 4: Calculate the largest multiple of LCM
Now, we multiply the LCM by 16:
\[ 612 × 16 = 9792 \]

### Step 5: Adjust for the remainder
Since we need a number that leaves a remainder of 1 when divided by 6, 9, 12, and 17, we add 1 to the largest multiple we found:
\[ 9792 + 1 = 9793 \]

### Conclusion
The greatest 4-digit number that leaves a remainder of 1 when divided by 6, 9, 12, and 17 is **9793**.

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