getDPA :Abundant, deficient, and perfect number classifications: Solution Submission

Contributors
#1

What is your hint or solution suggestion?

function sumProperDivisors(n) {
    let divArr = [];
    for(let i = 1; i < n; i++) {
       if(n%i == 0) {divArr.push(i);} }
    //console.log(`divArr currently holds ${divArr}.\n`);
    let divTot = divArr.reduce((el,tot) => { tot+= el; return tot;}, 0);
    //console.log(`divTot is ${divTot}.\n`);
    return divTot; }

function getDPA(n) {
let dCnt=0, pCnt=0, aCnt = 0, propDiv;
for(let j = 1; j <= n; j++) {
    propDiv = sumProperDivisors(j);
  if(propDiv < j) {dCnt++;} 
  if(propDiv == j) {pCnt++;} 
  if(propDiv > j) {aCnt++;} 
}
//console.log(`deficient: ${dCnt}, perfect: ${pCnt},`+
//`abundant: ${aCnt}.\n`);
  return [dCnt, pCnt, aCnt];
}

Challenge: Abundant, deficient and perfect number classifications

Link to the challenge:

#2

Why store then only to sum them? Why not sum as you go?

#3

How about

function sumProperDivisors(n) {
  let total = 0;
  // Check each possible divisor
  for (let i = 1; i < n; i++) {
    if (n % i === 0) {
        total += i;
    }
  }
  return total;
}

function getDPA(n) {
  let deficient = 0, perfect = 0, abundant = 0;

  for (let j = 1; j <= n; j++) {
    const propDiv = sumProperDivisors(j);
    // Classify the number
    if (propDiv < j) {
      deficient++;
    } else if (propDiv === j) {
      perfect++;
    } else {
      abundant++;
    }
  }

  return [deficient, perfect, abundant];
}
#4