I wrote a code with javascript for this problem :

"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."

but the result is false and i don't know why? can you help me guys

my code is :

function multipleSum(n){   
   var sum = 0;
   for(var i = 1; i<n; i++){
     var m3 = 3 * i;
     var m5 = 5 * i;    
     if(m3 < n  ){
       sum=sum+m3
     }
     if(m5 < n ){
       sum=sum+m5;
     }
     //if(m3 > n && m5 > n) {console.log(m3,m5,sum);break;}
   }
   return sum
}
console.log(multipleSum(1000)) //266333 but correct one is 233168 why?

I wrote a code with javascript for this problem :

"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."

but the result is false and i don't know why? can you help me guys

my code is :

function multipleSum(n){   
   var sum = 0;
   for(var i = 1; i<n; i++){
     var m3 = 3 * i;
     var m5 = 5 * i;    
     if(m3 < n  ){
       sum=sum+m3
     }
     if(m5 < n ){
       sum=sum+m5;
     }
     //if(m3 > n && m5 > n) {console.log(m3,m5,sum);break;}
   }
   return sum
}
console.log(multipleSum(1000)) //266333 but correct one is 233168 why?

• javascript

• 1 Hint: Why are you multiplying each number by 3 and 5? – charlietfl Commented Jun 24, 2018 at 11:48
• @charlietfl that's actually a very good approach now when I think about it. I just made changes in my answer to fit his solution. – kiddorails Commented Jun 24, 2018 at 11:57

4 Answers 4

Your logic is flawed. You should be iterating on each number (specified in range), and see if the modulus of the number with 3 or 5 is 0 or not. If modulus is zero, it means the number is divisible.

function multipleSum(n){
  var sum = 0;
  for(var i = 1; i<n; i++){
    if(i % 3 == 0 || i % 5 ==0){ // gives reminder of 0, divisible by either 3 or 5
      sum += i; // add in sum if that's the case.
    }
  }
  return sum
}
console.log(multipleSum(1000))

Edit: tried some time understanding why you went with multiply approach, I think you are gathering factors and want to break out early from the loop instead of iterating on entire collection. This should help you:

function multipleSum(n){
  var sum = 0;
  for(var i = 1; i<n; i++){
    var m3 = i * 3;
    var m5 = i * 5;
    if(m3 > n) break; // breaks early!!
    if(m3 < n) sum += m3
    if(m5 < n && m5 % 3 != 0) sum += m5; // make sure number is not divisible by 3, say m5 = 15, it will be captured as multiple of 3 anyway, we don't want duplicates.
  }
  return sum
}

console.log(multipleSum(1000))

Your logic is flawed in the way that, all the multiplications of 3 * 5 is doubled. Remember, you have:

3 * 1
5 * 1
3 * 2
3 * 3
5 * 2
3 * 4
3 * 5
5 * 3   // Here es the dupe.

I would do this in a different way.

1. Get all the multiples of 3 in an array.
2. Get all the multiples of 5 in an array.
3. Break the loop when both the multiplications are greater than n.
4. Merge both the arrays.
5. Remove the duplicates.
6. Add everything using the .reduce() function.

var num = 1000;
var m3 = [];
var m5 = [];

for (i = 0; i < num; i++) {
  if (i * 3 < num)
    m3.push(i * 3);
  if (i * 5 < num)
    m5.push(i * 5);
  if (i * 3 > num)
    break;
}

m35 = m3.concat(m5);

m35u = m35.filter(function(item, pos) {
  return m35.indexOf(item) == pos;
});

console.log(m35u.reduce((a, b) => a + b, 0));

I get 233168 as answer.

You can try this one liner (your home work: explain how this works ;):

console.log( 
  Array.from({length: 1000})
    .reduce( (p, n, i) => p + (i % 3 === 0 || i % 5 === 0 ? i : 0), 0 )
);

Try this, maybe answer you.Thank

const solution = (numb) => {
    const collectedNumb = [];
    const maxDividing = parseInt(numb / 3);
  
    for (let idx = 1; idx <= maxDividing; idx++) {
      const multipled3 = idx * 3;
      const multipled5 = idx * 5;
  
      multipled3 < numb && collectedNumb.push(multipled3);
      multipled5 < numb && collectedNumb.push(multipled5);
    }
    const uniqCollected = [...new Set(collectedNumb)].sort((a, b)=> a-b);
  
    console.log(uniqCollected);
  
    const reduced = uniqCollected.reduce((acc, numb) => acc + numb, 0);
    return reduced;
  };