Permutations and Combinations

Case I Let one digit be 1 or 7 , then sum of the other two digits be 2,5,8,11 or 14 or 3x+2 Therefore, the 2 digits can be (3,5),(5,9).
∴ The possible number are 8 .
Case II Let one digit be 3 or 9 the sum of other two digits can be multiple of 3 . The other 2 digit can be (1,5) or (7,5)∴ The possible numbers are 8 .
Case III Let one digit can be 5 , then sum of other digit can be 1,4,7,10,13 or 16 .
∴ The other 2 digit numbers can be (1,3),(1,9),(3,7),(7,6)
∴ The possible number can be 8 in ‌−number.
Hence, the required number are
8+8+8=24