for the set of all integers 1 to 6000 inclusive, there are obviously 6000/2=3000 elements that are divisible by 2. 6000/3=2000 elements that are divisible by 3 and 6000/5=1200 that are divisible by 5. The sum of these is 3000+2000+1200=6200. You must now subtract the numbers which we have counted multiple times. e.g. number 6 will be counted twice as it is divisible by 3 and 2. all the numbers divisible by both 2 and 3 will be multiples of 6, so there are 6000/6=1000 of these such numbers. all the numbers divisible by 2 and 5 will be multiples of 10, so there are 6000/10=600 of these such numbers. all the numbers divisible by 3 and 5 are multiples of 15, so there are 6000/15=400 of these such numbers. However the numbers divisible by all of 2,3 and 5 have been removed 3 times so must be added again. There are 6000/(2*3*5)=6000/30=200 numbers divisible by 2,3 and 5. Therefor the number of multiples of 2,3 or 5, in the integers 1-6000 is 3000+2000+1200-1000-600-400+200=4400. therefore, the number of integers that are not multiples is 6000-4400=1600.