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Car Wash at stake - If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23. Find the sum of all the multiples of 3 or 5 below 1000.

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DIS IS REAL.
You're real
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Other answers:

No just answers...
var sum = 0; for(var i = 1; i < 1000; i++) { if (i % 5 == 0 || i % 3 == 0) sum += i }
give me a link to your code or scratch paper
Baller!
I bequeath the car wash to darthsid.
Winning is mattfeury!
very honorable feury
He's got a cute face.
Side, clean up dat Rav 4 and get some spinning rims while you're at it
they're hipster now, and listen to some baumstein at 11
ok so apparently my attempt to calculate it manually is wrong, can you guys figure out what is wrong with it? 1. Multiples of 3: There are 3 multiples of 3 in the first 10 numbers, so there are 1000mod3 = 333 multiples of 3 lower than 1000. So their sum should be 3(1+2+3...+333) 2. Multiples of 5: There are 2 multiples of 5 in the first 10 numbers, so there are 200 multiples of 5 in 1000. Their sum is 5(1+2+3..+200) The total comes out to be wrong. What's wrong?
So I have to not count 1000, and I have to subtract the sum of common multiples (of 15). But it is still wrong
Sid, you are the Ramanujan of this riddle
MY LOGIC IS CORRECT! Only mistake was in the calcs.
Sid, you are the Rashomon of this riddle.
This is elementary mutual inclusion exclusion.
And this is actually project Euler problem #1 http://projecteuler.net/problem=1
i believe this should be in math section;) jk
You could just do the following for all number between 1 and 999: 1) find the sum of all multiples of 3 - call it Sum3 2) find the sum of all multiples of 5 - call it Sum5 3) find the sum of all multiples of 15 - call it Sum15 the your answer would be Sum = Sum3 + Sum5 - Sum15 I subtract Sum15 from this because any multiple of 15 will be included in both Sum3 AND Sum5.

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