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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.
 2 years ago
 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.
 2 years ago
 2 years ago

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mattfeuryBest ResponseYou've already chosen the best response.3
var sum = 0; for(var i = 1; i < 1000; i++) { if (i % 5 == 0  i % 3 == 0) sum += i }
 2 years ago

dpflanBest ResponseYou've already chosen the best response.0
give me a link to your code or scratch paper
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.3
I bequeath the car wash to darthsid.
 2 years ago

farmdawgnationBest ResponseYou've already chosen the best response.0
Winning is mattfeury!
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.3
He's got a cute face.
 2 years ago

dpflanBest ResponseYou've already chosen the best response.0
Side, clean up dat Rav 4 and get some spinning rims while you're at it
 2 years ago

dpflanBest ResponseYou've already chosen the best response.0
they're hipster now, and listen to some baumstein at 11
 2 years ago

darthsidBest ResponseYou've already chosen the best response.3
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?
 2 years ago

darthsidBest ResponseYou've already chosen the best response.3
So I have to not count 1000, and I have to subtract the sum of common multiples (of 15). But it is still wrong
 2 years ago

dpflanBest ResponseYou've already chosen the best response.0
Sid, you are the Ramanujan of this riddle
 2 years ago

darthsidBest ResponseYou've already chosen the best response.3
MY LOGIC IS CORRECT! Only mistake was in the calcs.
 2 years ago

mattfeuryBest ResponseYou've already chosen the best response.3
Sid, you are the Rashomon of this riddle.
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
This is elementary mutual inclusion exclusion.
 2 years ago

FoolForMathBest ResponseYou've already chosen the best response.0
And this is actually project Euler problem #1 http://projecteuler.net/problem=1
 2 years ago

karatechopperBest ResponseYou've already chosen the best response.0
i believe this should be in math section;) jk
 2 years ago

asnaseerBest ResponseYou've already chosen the best response.1
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.
 2 years ago
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