get all 3 digit number whose sum is 11 or we can say module is 1 in JS [closed]

I’m not sure what the second check is supposed to be (the modulus of something has to be 1). If that is supposed to be that the modulus of all digits of the number has to be 1, here is how you could do it:

var results = [];
for (let i = 1; i < 10; i++) {
  for (let j = 0; j < 10; j++) {
    for (let k = 0; k < 10; k++) {
      if (i + j + k === 11 && i % j % k === 1) {
        results.push(i.toString() + j.toString() + k.toString());
      }
    }
  }
}

console.log(results);

OR:

var results = [];
for (let i = 100; i < 1000; i++) {
  const [one, two, three] = i.toString().split('').map(i => +i);
  if (one + two + three === 11 && one % two % three === 1) {
    results.push(i.toString());
  }
}

console.log(results);

Basically just go through all the possible combination and do the checks on each number.

Leave a Comment