Using each of the digits 0, 1, 2, 3, 4 and 5 exactly once, how many 6-digit numbers divisible by 4 can be formed?

Last two digits must be a multiple of 4 with different digits: 04, 12, 20, 24, 32, 40, 52. Endings with 0 (04, 20, 40): remaining 4 digits in $4!=24$ ways each = 72. Endings without 0 (12, 24, 32, 52): 0 cannot lead, $4!-3!=18$ each = 72. Total $72+72=144$.