Explanation. Standard caveat: don’t look here if you are trying to do these yourself.

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.This one’s pretty straight forward, really, as one might hope being the first question. R’s built-in subsetting mechanism handles the extraction fairly nicely. I perhaps would have liked a way to do this without first definingFind the sum of all the multiples of 3 or 5 below 1000.

`x`

; though I suppose it could just be repeated in the last line.
[code language=“r”]

## check the worked solution

sum(c(3,5,6,9)) # 23

## values < 1000

x <- 1:999

## sum of x % 3 or x % 5

sum(x[x %% 3 == 0 | x %% 5 == 0]) # 233168

### CORRECT

[/code]