# Euler Problem 1: Multiples of 3 or 5

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## Euler Problem 1 Definition

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.

## Solution

There are four ways to solve this problem in R.

- Brute force loop through all numbers from 1 to 999 and test whether they are divisible by 3 or by 5 using the modulus function.
- Vector arithmetics.
- Sequences of 3 and 5, excluding duplicates (numbers divisible by 15).
- Using an arithmetic approach.

By the way, the problem definition on the Project Euler website is not consistent: the title mentions multiples of 3 AND 5, while the description asks for multiples of 3 OR 5.

# Solution 1 answer <- 0 for (i in 1:999) { if (i%%3 == 0 | i%%5 == 0) answer <- answer + i } # Solution 2 sum((1:999)[((1:999)%%3 == 0) | ((1:999)%%5 == 0)]) # Solution 3 sum(unique(c(seq(3, 999, 3), seq(5, 999, 5))))

The sum of an arithmetic progression , where *n* is the number of elements and *a _{1}* and

*a*are the lowest and highest value, is:

_{n}The numbers divisible by can be expressed as:

We can now easily calculate the sum of all divisors by combining the above progression with the formula for arithmetic progressions as expressed in the above code, where *m* is the divisor and *n<\i> the extent of the sequence.*

*p* is the highest number less than *n* divisible by *m*. In the case of 5, this number is 995.

Substitution gives:

# Solution 4 SumDivBy <- function(m, n) { p <- floor(n/m)*m # Round to multiple of n return (m*(p/m)*(1+(p/m))/2) } answer <- SumDivBy(3, 999) + SumDivBy(5, 999) - SumDivBy(15, 999)

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