Find the greatest 6-digit number which when devide by 6,12,18, 24 leaves remainder 5 in each case?

999941

The first step here is to find the numbers into which 6,12,18,24 divide evenly into, that is, their lowest common multiple (hint: 72). The largest multiple of 72 not exceeding six digits is 72 x13888 = 999936. Add five to this and you will get the greatest six digits into which they all divide leaving 5 remaining, so 999936 +5 = 999941.