# How many 3-digit positive integers are divisible by 6 but not by 8?

a) 130

b) 225

c) 113

d) 150

e) 260

We calculate all the integers in the given range which are divisible by 6 and then subtract those which are divisible by both 6 and 8. The latter condition is the same as "divisible by 24".

The first integer in [100,999] divisible by 6 is 102. The last one is 996. So the number of integers divisible by 6 is (996-102) / 6 + 1 = 150

Similarly, the first integer divisible by 24 in the interval is 120. The last is 984. So there are (984-120)/24 + 1 = 37 integers divisible by 24 in the interval.

The number of integers divisible by 6 but not by 8 (i.e. not by 24) is

150 - 37 = 113

Ans: c)

Jul 20 at 13:58

the 3 digit positive integers begin with 100, and end with 999. That's a total of 900 possible numbers.

of these 900 numbers, the first one divisible by 6 is 102. That leaves 988 possible numbers divisible by 6. the last number divisible by 6 is 996, so we have whittled down the range of numbers divisible by 6 to 993.

notice that 996 = 6*166 and 102 = 6*17. this means we have (166-17) + 1 = 150 multiples of 6 in this range.

so how many of these 150 multiples of 6 are also multiples of 8? Since the least common multiple of 6 and 8 is 24, once we find the first one, they will occur every 24th integer. Checking in turn, 102, 108, 114 and 120, we find the least 3 digit multiple of 6 also divisible by 8 is 120.

996-120 = 876, and 876/24 = 36.5. Thus we would expect to find 36 more multiples of 24 (divisible by 6 AND by 8) between 120 and 996. This brings our total (including 120) to 37.

150 - 37 = 113, the correct answer is c).

Jul 20 at 17:44