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
By hand
