Find the greatest number of five digits which when divided by 3, 5, 8, 12 have 2 as remainder :

Options

A. 99999
B. 99958
C. 99960
D. 99962

Answer

To solve the problem of finding the greatest five-digit number that leaves a remainder of 2 when divided by 3, 5, 8, and 12, 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 3, 5, 8, and 12.

- The prime factorization of each number is:
- 3 = 3^1
- 5 = 5^1
- 8 = 2^3
- 12 = 2^2 * 3^1

- The LCM is found by taking the highest power of each prime:
- For 2: max(3, 2) = 3 → 2^3
- For 3: max(1, 1) = 1 → 3^1
- For 5: max(1) = 1 → 5^1

Thus, the LCM is:
\[ \text{LCM} = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120 \]

### Step 2: Identify the greatest five-digit number
The greatest five-digit number is 99999.

### Step 3: Determine the largest multiple of LCM less than or equal to 99999
To find the largest multiple of 120 that is less than or equal to 99999, we divide 99999 by 120 and take the integer part of the result:

\[ 99999 \div 120 = 833.325 \]

Taking the integer part, we get 833. Now, we multiply back to find the largest multiple of 120:

\[ 833 \times 120 = 99960 \]

### Step 4: Adjust for the remainder
Since we need a number that leaves a remainder of 2 when divided by 3, 5, 8, and 12, we add 2 to the largest multiple of 120:

\[ 99960 + 2 = 99962 \]

### Conclusion
The greatest five-digit number that leaves a remainder of 2 when divided by 3, 5, 8, and 12 is:

\[ \boxed{99962} \]