The sum of the first two is divisible by the second, but not by the first.
The sum of the first three is divisible by the third, but not by the first or the second.
The sum of the first four is divisible by the fourth, but not by the first, second or third.
The sum of the first five is divisible by the fifth, but not by the first, second, third or fourth.

What are the five integers, in order?

1.           Bear in mind that the final number can be 1, which divides every number, but is not divisible by any other. No other number can be 1.

2.           Since the sum of the first two integers is divisible by the second but not by the first, the first number must be a multiple of the second.

3.           None of the numbers can be 2, as 2 divides all even numbers, and at least two of the first four numbers must be even.

4.           Since all the numbers are less than 10, the second must be either 3 or 4.

5.           There are now 29 solutions where b=3 and 16 where b=4, when a= {6,9} or a=8. If a=6, c={1,3,9} which will not work. If a=9, c={1,2,3,4,6} and all except 4 do not work (c cannot be 1 or 2, and 3 divides 6). Assume that c=4, then d={1,2,4,8} and we can discard {1,4,8} leaving 2, but 2 cannot work.

6.  Assume that a=8; then c={1,2,3,4,6}, and we can discard those solutions where d=1, 4 or 8, or d=c.

7.  If c=3,d=5, or c=6,d=2 then 4 divides 20, which does not work. b=4, c=4 and b=6,c=6 do not work, leaving {8,4,6,3,1} since the other available alternatives are divisible either by 6 or 3.

8. So the solution is {8,4,6,3,1}; 1 divides 22=2*11, which has no common divisors with any of the others, 3 divides 21 7 times, 6 divides 18 3 times, 4 divides 12 3 times.

I believe this problem was in New Scientist a few months ago, this probably should be acknowledged.