Project Euler with Julia
The first problem is very simple:
Multiples of 3 and 5
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 multip les is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
One easy-to-code way is by dealing with arrays:
julia> sum( filter( x->((x%3==0) | (x%5==0)), [1:999] ) ) 233168
A much better way is to obtain it via summation formulas: Note that the sum of all multiples of three between one and 999 is given by
Similarly, the sum of all multiples of five between one and 995 is given by
Finally, we need to subtract the sum of the multiples of both three and five, since they have been counted twice. This sum is
The final sum is then,