None.
All multiples of 536, which is an infinite number.
536, 1072, 1608
268, 536, 804 and keep adding 268 forever.
The total number of integers that are multiples of both 67 and 8 is infinite. The first few are these: 536, 1072, 1608, 2144 . . .
525 and 25
102
The multiples of 268 between 1 and 10 are: 1 x 268 = 268 2 x 268 = 536 3 x 268 = 804 4 x 268 = 1072 5 x 268 = 1340 6 x 268 = 1608 7 x 268 = 1876 8 x 268 = 2144 9 x 268 = 2412 10 x 268 = 2680
134, 268, 402, 536, 670, 804, 938, 1072, 1206, 1340, 1474, . . .
134, 268, 402, 536, 670, 804, 938, 1072, 1206, 1340.
67....................................................................................................................................................................................................................................................... . ....................... . . . ..................... . . . . . . . ..........
23.3043
There are infinitely many solutions for that. Choose any number for your first number. Then subtract 536 minus that number, to get the other one.