find the greatest number of five digits which when divided by 3 5 8 12

The greatest five-digit number that satisfies this condition is 99962.

Step-by-step reasoning

1. Let the required number be NNN.
   When NNN is divided by 3, 5, 8, or 12, the remainder is 2, so N−2N-2N−2 is exactly divisible by all of them.

2. That means N−2N-2N−2 must be a multiple of the LCM of 3, 5, 8, and 12.
   • 3=313=3^13=31
   • 5=515=5^15=51
   • 8=238=2^38=23
   • 12=22×3112=2^2\times 3^112=22×31
     Taking highest powers: 23,31,512^3,3^1,5^123,31,51, so
     LCM=23×3×5=8×3×5=120\text{LCM}=2^3\times 3\times 5=8\times 3\times 5=120LCM=23×3×5=8×3×5=120.

3. So N−2N-2N−2 is a multiple of 120.
   The greatest five-digit number is 99999, so we want the largest multiple of 120 that is ≤ 99999.

4. Divide to find that multiple:
   99999÷120=833.32599999\div 120=833.32599999÷120=833.325, so take the integer part 833.
   Then 833×120=99960833\times 120=99960833×120=99960, which is the greatest five-digit number divisible by 120.

5. Since N−2=99960N-2=99960N−2=99960, add back 2:
   N=99960+2=99962N=99960+2=99962N=99960+2=99962.

So, the required greatest five-digit number which when divided by 3, 5, 8, and 12 leaves remainder 2 is 99962.