The key to this is the line CAAC. That number must be a multiple of 11 (sum of digits in even-numbered positions = sum of digits in odd-numbered positions). Since D is obviously not a multiple of 11, it follows that ABC, and hence also IDBDD, must be multiples of 11.

Assuming that the letters A to I stand for the numbers 1 to 9 in some order, you should be able to see fairly soon that D=6. Thus the bottom line becomes I6B66, which can only be a multiple of 11 if I+B = 6 or 17. You can rule out 17 quite easily, and then you soon find the remaining values.