Problem 1: Multiples of 3 and 5 Why my function returns false result for any number besides 10?

JavaScript
1

Tell us what’s happening:

Look at console logs, you’ll see that the array is supposed to be right since it pushes numbers under the argument"number"

Your code so far


function multiplesOf3and5(number) {
    var sum = [];
    var multi = 0;
    for(var i = 0; multi*3 < number;multi++) {
      i = multi*3;
      sum.push(i);
    }
    console.log(sum);
    multi = 0;
    for(var j = 0; multi*5 < number;multi++) {
      j = multi*5;
      sum.push(j);
    }
    console.log(sum);
    return sum.reduce((a,b) => a + b);
}

multiplesOf3and5(49);

Your browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:60.0) Gecko/20100101 Firefox/60.0.

Link to the challenge:
https://learn.freecodecamp.org/coding-interview-prep/project-euler/problem-1-multiples-of-3-and-5/

2

In your example call above multiiplesOf3and5(49), you are double counting some numbers. For example, you are counting 15, 30, and 45 twice.

1 Like
3

Yeah, that was because there were some numbers divisible(x % y = 0) by 5 or 3. Now I added an if condition to prevent that from happening:

function multiplesOf3and5(number) {
    var sum = [];
    var multi = 0;
    for(var i = 0; multi*3 < number;multi++) {
      sum.push(multi*3);
    }
    multi = 0;
    for(var j = 0; multi*5 < number;multi++) {
      if (multi*5 % 3 > 0) {
      sum.push(multi*5);
      }
    }
    return sum.reduce((a,b) => a + b,0);
}

multiplesOf3and5(10);

Thank you for the help!

4

I have been playing more with the code and seen a much better solution:

function multiplesOf3and5(number) {
    var sum = 0; // sum for multiples of 3 and 5;
    for(var i = 0; i < number;i++) { // a loop to increase i everytime
      if (i % 3 === 0 || i % 5 === 0) { //this condition is important because everytime i increases, this condition checks if it's a multiple(divisible by) for either 3 or 5. 
       sum += i; //increases sum with i that is divisible with 3 and 5
      }
    }
    return sum; //returns sum of numbers below "number" that are multiples of 3 and 5
}

multiplesOf3and5(10);

It’s shorter and cleaner.

5