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Three digits numbers which are divisible by 6 are 102, 108, 114....996This will form an arithmetic progression havingFirst term (a) = 102Common difference (d) = 6 and last term (a n ) = 996We know, a n = a + (n - 1) X dOr, 996 = 102 + (n - 1) X 6Or, (n - 1) = {(996 - 102) ÷ 6Or, (n - 1) = 149Or, 'n' = 150So, number of 3 digit numbers which are divisible by 6 = 150Similarly, number of 3-digit numbers divisible by 9 = {(999 - 108) /9} + 1 = 100Number of 3 digits numbers divisible by both 6 and 9 are the numbers divisible by LCM of (6, 9) , i.e., 18 = {(990 - 108) /18} + 1 = 50So, the required numbers are = 150 + 100 - 50 = 200
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