I’ll try to solve two or three PE problems per week and post here, along with an explanation of how I reached the result. So, let’s start with… 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.
OK. So let’s use Python on this one. There are many different ways we could do this – I chose something more “brute”.
Okay. So, what are we doing? First, we tell Python the variable number stores the value of
0. In other words, the variable is there but holds nothing. Then, we say this: while i stays between 1 and 1000, do this – and then we pass a condition. In line 3 – where the condition is passed – there is the modulo operator - it finds the remainder of a division. So, for a number to be multiple of 3 or 5, it needs to have a remainder of 0 when we divide it by 3 or 5, right? That’s what we’re doing. Now that we have passed this condition, we can simply sum i. Remember: this will only happen when i has a remainder of 0 when divided by 3 or 5. After we increment number with i and all the loops have been done, we print number.
There are many other ways we could have done that. Instead of number = number + i, it is possible to simply write number += i. It is also possible to do this with arithmetic progressions.
Okay, that’s it!