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dpflan

  • 2 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.

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  1. farmdawgnation
    • 2 years ago
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    DIS IS REAL.

  2. dpflan
    • 2 years ago
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    You're real

  3. mattfeury
    • 2 years ago
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    233168

  4. dpflan
    • 2 years ago
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    No just answers...

  5. mattfeury
    • 2 years ago
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    var sum = 0; for(var i = 1; i < 1000; i++) { if (i % 5 == 0 || i % 3 == 0) sum += i }

  6. dpflan
    • 2 years ago
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    give me a link to your code or scratch paper

  7. dpflan
    • 2 years ago
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    Baller!

  8. mattfeury
    • 2 years ago
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    I bequeath the car wash to darthsid.

  9. farmdawgnation
    • 2 years ago
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    Winning is mattfeury!

  10. dpflan
    • 2 years ago
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    very honorable feury

  11. mattfeury
    • 2 years ago
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    He's got a cute face.

  12. dpflan
    • 2 years ago
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    Side, clean up dat Rav 4 and get some spinning rims while you're at it

  13. dpflan
    • 2 years ago
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    they're hipster now, and listen to some baumstein at 11

  14. darthsid
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    Sid, you are the Ramanujan of this riddle

  17. darthsid
    • 2 years ago
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    MY LOGIC IS CORRECT! Only mistake was in the calcs.

  18. mattfeury
    • 2 years ago
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    Sid, you are the Rashomon of this riddle.

  19. FoolForMath
    • 2 years ago
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    This is elementary mutual inclusion exclusion.

  20. FoolForMath
    • 2 years ago
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    And this is actually project Euler problem #1 http://projecteuler.net/problem=1

  21. karatechopper
    • 2 years ago
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    i believe this should be in math section;) jk

  22. asnaseer
    • 2 years ago
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    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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