For how many ordered triplets (a, b, c) of positive integers, each less than 10 is the product
abc divisible by 20?

5 always has to be one of the numbers, so this is equivalent to looking for two numbers less than 9 whose product is divisible by 4. I did it by hand, and seem to have found 21 (same as Ed). I found it easier to order decreasing. My list

98, 96, 94

88, 87, 86, 85, 84, 83, 82, 81

74

66, 64, 62

54

44, 43, 42, 41

22