Finding Integers Divisible by 2, 3 or 5 - Solve Here!

[Physics is Phun]

We have a question that asks to find the number of integers between 1 and 100 that are divisible by 2,3 or 5.
So i use the sum rule

let E,F and G be the the integers. so,
n(EUFUG) = n(E) + n(F) + n(G) - n(EnF) - n(EnG) - n(FnG)
but this doesn't work. it gives me the answer of 70 but the actual answer is 74. Since I know the answer is 74 this one doesn't matter but the next question asks to find the number of integers between 1 and 100000. So it's obvious that I can't count them.
just to make sure I did the above part right
n(EUFUG) would be 100/2 + 100/3 + 100/5 - 100/6 - 100/10 - 100/15 correct?
I hope someone can figure out where I've gone wrong.

 

[vincentchan]

you forget to add the number divisible by both 2,3,5, in your notation, it is n(EnFnG)
2*3*5=30, so, you got to add 3 in your final answer, one more thing,n(E) + n(F) + n(G) - n(EnF) - n(EnG) - n(FnG)=71 instead of 70, you made a arithmatic mistake, also

 

[ShawnD]

just to make sure I did the above part right
n(EUFUG) would be 100/2 + 100/3 + 100/5 - 100/6 - 100/10 - 100/15

I'm not familiar with what you are doing, but wouldn't you also need to subtract 100/30? 100/6 would be the ones common between 2 and 3, 100/10 would be the ones common between 2 and 5, 100/15 would be the ones common between 3 and 5, and 100/30 would be the ones common between 2, 3, and 5.
Of course it would make you even farther from the "correct" answer, but I'm just saying.

 

[Parth Dave]

the ones that are common to 2, 3, and 5 were taken care of when you took account for the other combinations. Not only that, but it was subtracted twice. Thats why you have to add it again.

 

[Physics is Phun]

I still don't get it. when using the sum rule, you add the parts together, the subract the overlaping sections. I don't see why n(EnFnG) needs to be added. and thanks for the arithemtic point out. I was adding the exact values and not rounding each section which actually makes the answer more wrong (unlike most math problems)

 

[ShawnD]

the ones that are common to 2, 3, and 5 were taken care of when you took account for the other combinations. Not only that, but it was subtracted twice. Thats why you have to add it again.

I was going to challenge you on that, but I guess you're right. Try it with 1, 2, and 3 between 1 and 10. The real answer is obviously 10, so it's easy to check the method.
Here are each of em:
10/1 = 10, 10/2 = 5, 10/3 = 3. Sum: 18
The common between 2:
10/2 = 5, 10/3 = 3, 10/6 = 1. Sum: 9
The common between 3:
10 / 6 = 1
Subtract the ones common between 2 from from the single ones:
18 - 9 = 9. We're trying to get to 10, so add the one common between all 3 to get 10

 

[Physics is Phun]

ok I get it now. I just had to reread ur post parth dave.
thanks
:)