Project Euler 1

Problem 1

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.

Solve：

clear
clc
a = 0;
b = 0;
c = 0;
for i = (1:1:999)
if mod(i,3) == 0
a = a+i; %3的倍数和
end
if mod(i,5) == 0
b = b+i; %5的倍数和
end
if mod(i,15) == 0
c = c+i; %既是3又是5的倍数和
end
end
n = a+b-c; 
fprintf('The answer is %.0d\n',n)