Find out the greatest number of four digits, which when divided by 6, 9, 12, 15 and 18 leaves the remainder 2 in each case.

Options

A. 9900
B. 9901
C. 9902
D. 9990

Answer

To solve the problem of finding the greatest four-digit number that leaves a remainder of 2 when divided by 6, 9, 12, 15, and 18, we can follow these steps:

### Step 1: Find the LCM of the Divisors
We need to find the least common multiple (LCM) of the numbers 6, 9, 12, 15, and 18.

- **Prime factorization:**
- 6 = 2 × 3
- 9 = 3²
- 12 = 2² × 3
- 15 = 3 × 5
- 18 = 2 × 3²

- **Taking the highest powers of all prime factors:**
- For 2: The highest power is 2² (from 12).
- For 3: The highest power is 3² (from 9).
- For 5: The highest power is 5¹ (from 15).

- **Calculating the LCM:**
\[ \text{LCM} = 2² × 3² × 5¹ = 4 × 9 × 5 = 180 \]

### Step 2: Identify the Largest Four-Digit Number
The largest four-digit number is 9999.

### Step 3: Adjust for the Remainder
Since we want a number that leaves a remainder of 2 when divided by 6, 9, 12, 15, and 18, we need to subtract the remainder (2) from the largest four-digit number:

\[ 9999 - 2 = 9997 \]

### Step 4: Find the Largest Number Divisible by the LCM
Now we need to find the largest number less than or equal to 9997 that is divisible by the LCM (180). We can do this by dividing 9997 by 180 and taking the floor of the result:

\[ 9997 \div 180 \approx 55.52 \]

Taking the floor gives us 55. Now, we multiply back by 180 to find the largest multiple of 180:

\[ 55 × 180 = 9900 \]

### Step 5: Add the Remainder
Finally, to ensure that this number leaves a remainder of 2, we add 2:

\[ 9900 + 2 = 9902 \]

### Conclusion
Thus, the greatest four-digit number that leaves a remainder of 2 when divided by 6, 9, 12, 15, and 18 is **9902**.

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