How print out numbers for multiples of 3 that divide without a remainder numbers?

Question

How can I print out numbers for multiples of 3 that divide without a remainder numbers?

I am attempting to get 3 , 6 , and 9 in array but only 1 prints out, my syntax could be wrong.

 var numbers = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]; for (var i = 0; i % 3 === 0; i++) { console.log (numbers[i]); }

Answer 1

Two things:

1. When this i % 3 === 0 check fails your loop stops.

2. You should check whether the elements( numbers[i] ) of your array are divisible by 3, not the indices( i ).

 var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]; for (var i = 0; i < numbers.length; i++) { if (numbers[i] % 3 === 0) { console.log(numbers[i]); } }

Answer 2

You should understand more about for loop. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/for

Code Some think like that:

var numbers = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ];

for (var  i = 0; i <= 10 ; i++) {
  if(numbers[i] % 3 === 0){
    console.log (numbers[i]);
  }
}

Answer 3

var numbers = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]; for (var i = 0; i <= 10; i++){ if( i % 3 === 0 ) { console.log(numbers[i]); } }

Answer 4

var numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10];

const res = numbers.forEach((el) => {
  if (el % 3 === 0) {
    console.log(el);
  }
});

just one more way to do it using forEach. LOGIC - If the number is divisible by three then it is a multiple of 3. Try testing to see if when the number is divided by 3 there is a remainder using the modulus operator