Euler Project #1 in R

Question

Problem

Find the sum of all numbers below 1000 that can be divisible by 3 or 5

One solution I created:

x <- c(1:999)
values <- x[x %% 3 == 0 | x %% 5 == 0]
sum(values

Second solution I can't get to work and need help with. I've pasted it below. I'm trying to use a loop (here, I use while() and after this I'll try for()). I am still struggling with keeping references to indexes (locations in a vector) separate from values/observations within vectors. Loops seem to make it more challenging for me to distinguish the two.

Why does this not produce the answer to Euler #1?

x <- 0
i <- 1
while (i < 100) {
  if (i %% 3 == 0 | i %% 5 == 0) {
    x[i] <- c(x, i)
  }  
  i <- i + 1
}
sum(x)

And in words, line by line this is what I understand is happening:

1. x gets value 0
2. i gets value 1
3. while object i's value (not the index #) is < 1000
4. if is divisible by 3 or 5
5. add that number i to the vector x
6. add 1 to i in order (in order to keep the loop going to defined limit of 1e3
7. sum all items in vector x

I am guessing x[i] <- c(x, i) is not the right way to add an element to vector x. How do I fix this and what else is not accurate?

Answer 1

A very efficient approach is the following:

div_sum <- function(x, n) {
  # calculates the double of the sum of all integers from 1 to n 
  # that are divisible by x
  max_num <- n %/% x
  (x * (max_num + 1) * max_num)      
}

n <- 999
a <- 3
b <- 5

(div_sum(a, n) + div_sum(b, n) - div_sum(a * b, n)) / 2

---

In contrast, a very short code is the following:

x=1:999
sum(x[!x%%3|!x%%5])

Answer 2

Here is a shortcut that performs this sum, which is probably more in the spirit of the problem:

3*(333*334/2) + 5*(199*200/2) - 15*(66*67/2)
## [1] 233168

Here's why this works:

In the set of integers [1,999] there are:

333 values that are divisible by 3. Their sum is 3*sum(1:333) or 3*(333*334/2).

199 values that are divisible by 5. Their sum is 5*sum(1:199) or 5*(199*200/2).

Adding these up gives a number that is too high by their intersection, which are the values that are divisible by 15. There are 66 such values, and their sum is 15*(1:66) or 15*(66*67/2)

As a function of N, this can be written:

f <- function(N) {
  threes <- floor(N/3)
  fives  <- floor(N/5)
  fifteens <- floor(N/15)

  3*(threes*(threes+1)/2) +  5*(fives*(fives+1)/2) - 15*(fifteens*(fifteens+1)/2)
}

Giving:

f(999)
## [1] 233168
f(99)
## [1] 2318

Answer 3

And another way:

x <- 1:999
sum(which(x%%5==0 | x%%3==0))
# [1] 233168