Hi there,

I need help with the following question.

How many integers are there between 100 and 800 divisable by 6 or 7 or both?

Okay lets start with the 6, there are two ways to do it
1. Anything divisible by 3, add up the numbers and divide by three. 102 is the lowest number after 100 that is divisible by three, 798 is the lowest number before 800 divisible by 3. Subtract these numbers and you get 696. Divide them by 3 and you get 232, divide that by 2 and you get 116. This works out because 6 is twice as much as three, so half of the integers from 3 will be divisible by 6.
2. 100/6 = 16 R2 So 102/6 = 17
800/6 = 133 R4 So 6-4 =2 and that is how far 800 is away from a multiple of 6. 798/6 = 133
798-102 = 696
696/6 = 116
So there are 116 integers between 100 and 800 divisible by 6


Now lets do 7
100/7 14 R2 so when 7 goes into 100 there is a remainder of 2 therefore 7-2 = 5, and 5 = 100 =105 the first multiple of 7 after 100. 105/7 = 15
800/7 = 114 R2 so the highest multiple in the range of 100 to 800 for 7 is 798.
So 798 - 105 = 693
693/7 = 99
So there are 99 integers between 100 and 800 divisible by 7

The last one is a bit tricky, but this is how I would do it. They want to know how many integers are divisible by both. All that means is add the amount of intergers they both have separately and subtract by the amount that they both share. You must subtract because you end up adding those numbers twice. Think about two circles that overlap. If you want to know the area that both circles cover, you add both areas separately, and then you subract the center area one time.
Use 42, because 7*6 = 42
100/42 = 2 R16 So 126 is the first multiple of 42
126/42 = 3
800/42 = 19 R2 So 798 is the last multiple of 42 in that range
798/42 = 19
798-126 = 672
672/42 = 16
So now that you know that 16 multiples between 100 and 800 are shared by 6 and 7, so you add 116 + 99 - 16= 199
So there are 199 integers between 100 and 800 divisible by both 6 and 7