all 3 digit number divisible by 6:
102 108 114 120 126 132 138 144...978 984 990 996
How many of those are there ?
(996-102)/6 + 1 = 150
or: An AG with a=102, d = 6, term(n) = 996
term(n) = a+(n-1)d
996 = 102 + 6(n-1)
6n-6 = 894
6n = 900
n = 150
Of those which ARE divisible by 9:
108 126 144 ... 990
This is another AG, where a = 108, d = 18, term(n) = 990
How many of those?
990 = 108 + 18(n-1)
882 = 18n - 18
18n = 900
n = 50
of those NOT divisible by 9 = 150 - 50 = 100
How many three-digit counting numbers
are exactly divisible by 6 but not also exactly divisible by 9?
2 answers
But why isn’t it 50????
900 / 6 = 150
900 / 9 = 100
150 - 100 = 50
900 / 6 = 150
900 / 9 = 100
150 - 100 = 50