Question

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.

Answer 1

DIS IS REAL.

Answer 2

You're real

Answer 3

233168

Answer 4

No just answers...

Answer 5

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

Answer 6

give me a link to your code or scratch paper

Answer 7

Baller!

Answer 8

I bequeath the car wash to darthsid.

Answer 9

Winning is mattfeury!

Answer 10

very honorable feury

Answer 11

He's got a cute face.

Answer 12

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

Answer 13

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

Answer 14

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?

Answer 15

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

Answer 16

Sid, you are the Ramanujan of this riddle

Answer 17

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

Answer 18

Sid, you are the Rashomon of this riddle.

Answer 19

This is elementary mutual inclusion exclusion.

Answer 20

And this is actually project Euler problem #1

http://projecteuler.net/problem=1

Answer 21

i believe this should be in math section;) jk

Answer 22

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.