## dpflan 3 years ago 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.

1. farmdawgnation

DIS IS REAL.

2. dpflan

You're real

3. mattfeury

233168

4. dpflan

5. mattfeury

var sum = 0; for(var i = 1; i < 1000; i++) { if (i % 5 == 0 || i % 3 == 0) sum += i }

6. dpflan

7. dpflan

Baller!

8. mattfeury

I bequeath the car wash to darthsid.

9. farmdawgnation

Winning is mattfeury!

10. dpflan

very honorable feury

11. mattfeury

He's got a cute face.

12. dpflan

Side, clean up dat Rav 4 and get some spinning rims while you're at it

13. dpflan

they're hipster now, and listen to some baumstein at 11

14. darthsid

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?

15. darthsid

So I have to not count 1000, and I have to subtract the sum of common multiples (of 15). But it is still wrong

16. dpflan

Sid, you are the Ramanujan of this riddle

17. darthsid

MY LOGIC IS CORRECT! Only mistake was in the calcs.

18. mattfeury

Sid, you are the Rashomon of this riddle.

19. FoolForMath

This is elementary mutual inclusion exclusion.

20. FoolForMath

And this is actually project Euler problem #1 http://projecteuler.net/problem=1

21. karatechopper

i believe this should be in math section;) jk

22. asnaseer

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.