How to calculate smallest number with certain number of divisors?

Question

From Project Euler problem 500

The number of divisors of 120 is 16. In fact 120 is the smallest number having 16 divisors.

Find the smallest number with 2**500500 divisors. Give your answer modulo 500500507.

It's simple enough to count the divisors of n, eg. in Python len([i for i in range(1,n+1) if n % i == 0]). This is O(n).

I tried brute force search and found the smallest number with 32 divisors is 840, but it's much too slow for the problem above. From the inequality count_divisors(n) <= n, the number is going to be massive.

What do you think? Any ideas? How to calculate smallest number with certain number of divisors?

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Edit: I don't think this is a duplicate. This question is more specific, it concerns a particular class of much larger numbers. The other question asks generally. Its answers aren't applicable to this problem, it's a different magnitude.

Answer 1

You should use the formula for the number of divisors of integer n:

d(n) = (a₁+1)(a₂+1)...(a_k+1)

where

n = p₁^a₁ * p₂^a₂ *p₃^a₃ *...*p_k^a_k

is a unique representation of every integer through powers of its prime divisors. This is a well-known formula, but if one wonders how to get it, note that d divides n if and only if d is of the form p₁^x₁ * p₂^x₂ *p₃^x₃ *...*p_k^x_k, where each of x_i is between 0 and a_i, so there are a_i + 1 possibilities for choosing each of x_i. Now just apply the product rule and you get the desired formula.

For fixed d(n) (as in your case), the minimum value of n is obviously obtained by carefully selecting powers of existing primes, or by adding new primes. Let's work through this simple example, 16:

d(x) = (a₁+1)(a₂+1)...(a_k+1) = 16 = 2⁴.

This means that you have at most 4 different primes, therefore:

x = 2^a₁ * 3^a₂ *5^a₃ * 7^a₄

where a_i >= 0. The question is now - in order to get minimum value of x, is it better to increase the powers of 2 (i.e., increment a₁), or to use 7 (i.e. take a₄=1 instead of a₄=0)? Well, it's simple to check, 2*3*5*7 > 2³ * 3 * 5 = 120, that's why the 120 is answer in this case.

How to generalize this approach? You should create min-heap where you'll put the powers of primes, taking care that the number of divisors reaches the specified value. In case of 16, this min-heap would contain numbers 2, 3, 5, 7, 2², 3², 2⁴ etc. Why? Because 16 = 2⁴, so each of (a_i+1) has to divide 16, i.e. it has to be power of 2. Every time when you add new power, it should increase the left hand side (i.e., the variable d(x)) by power of 2, since your final goal is to find the smallest number with 2⁵⁰⁰⁵⁰⁰ divisors. Heap is initialized with first k primes (in the problem statement, k = 500500), and in each step, when you pop p^x from the heap, p^2x is returned and result is multiplied by p^x. Step-by-step solution for d(x) = 16 = 2⁴:

Step    Heap    d(x)    x
==========================
0      2,3,5,7    1     1
1      3,4,5,7    2     2
2      4,5,7,9    4     6
3      5,7,9,16   8     24
4      7,9,16,25  16    120

HTH.

Answer 2

Here is the high level gist of my Javascript - where factorCount represents the number of divisors:

• Find the prime factor decomposition of the factorCount
• Generate every unique combination of these prime factors
• For each combination, extract these combination values from the original prime factor array and add one value to this array that is the extracted values multiplied together. Then sort elements in descending order.
• For each array created by the previous step, check which yields the minimal number when computing 2^(b1-1)*3^(b2-1)5^(b3-1)...
• This minimum number computed is the smallest number with factorCount number of divisors

Here's a high level code breakdown of my JavaScript functions:

var primeFactors = findPrimeFactors(factorCount);
var primeFactorCombinations = removeDuplicateArrays(generateCombinations(primeFactors, 1));
var combinedFactorCandidates = generateCombinedFactorCombinations(primeFactors, primeFactorCombinations);
var smallestNumberWithFactorCount = determineMinimumCobination(combinedFactorCandidates);

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And here's the full sha-bang:

function smallestNumberByFactorCount(factorCount) {

  function isPrime(primeCandidate) {
    var p = 2;
    var top = Math.floor(Math.sqrt(primeCandidate));
    while(p<=top){
      if(primeCandidate%p === 0){ return false; }
      p++;
    }
    return true;
  }

  function findPrimeAfter(currentPrime) {
    var nextPrimeCandidate = currentPrime + 1
    while(true) {
      if(isPrime(nextPrimeCandidate)){
        return nextPrimeCandidate;
      } else {
        nextPrimeCandidate++;
      }
    }
  }

  function findPrimeFactors(factorParent) {
    var primeFactors = [];
    var primeFactorCandidate = 2;
    while(factorParent !== 1){
      while(factorParent % primeFactorCandidate === 0 && factorParent !== 1 ){
        primeFactors.push(primeFactorCandidate);
        factorParent /= primeFactorCandidate;
      }
      primeFactorCandidate = findPrimeAfter(primeFactorCandidate);
    }
    return primeFactors;
  }

  function sortArrayByValue(a,b){
    return a-b;
  }

  function clone3DArray(arrayOfArrays) {
    var cloneArray = arrayOfArrays.map(function(arr) {
      return arr.slice();
    });
    return cloneArray;
  }

  function doesArrayOfArraysContainArray(arrayOfArrays, array){
    var aOA = clone3DArray(arrayOfArrays);
    var a = array.slice(0);
    for(let i=0; i<aOA.length; i++){
      if(aOA[i].sort().join(',') === a.sort().join(',')){
        return true;
      }
    }
    return false;
  }

  function removeDuplicateArrays(combinations) {
    var uniqueCombinations = []
    for(let c=0; c<combinations.length; c++){
      if(!doesArrayOfArraysContainArray(uniqueCombinations, combinations[c])){
        uniqueCombinations[uniqueCombinations.length] = combinations[c];
      }
    }
    return uniqueCombinations;
  }

  function generateCombinations(parentArray, minComboLength) {
    var generate = function(n, src, got, combinations) {
      if(n === 0){
        if(got.length > 0){
          combinations[combinations.length] = got;
        }
        return;
      }
      for (let j=0; j<src.length; j++){
        generate(n - 1, src.slice(j + 1), got.concat([src[j]]), combinations);
      }
      return;
    }
    var combinations = [];
    for(let i=minComboLength; i<parentArray.length; i++){
      generate(i, parentArray, [], combinations);
    }
    combinations.push(parentArray);
    return combinations;
  }

  function generateCombinedFactorCombinations(primeFactors, primeFactorCombinations) {
    var candidates = [];
    for(let p=0; p<primeFactorCombinations.length; p++){
      var product = 1;
      var primeFactorsCopy = primeFactors.slice(0);
      for(let q=0; q<primeFactorCombinations[p].length; q++){
        product *= primeFactorCombinations[p][q];
        primeFactorsCopy.splice(primeFactorsCopy.indexOf(primeFactorCombinations[p][q]), 1);
      }
      primeFactorsCopy.push(product);
      candidates[candidates.length] = primeFactorsCopy.sort(sortArrayByValue).reverse();
    }
    return candidates;
  }

  function determineMinimumCobination (candidates){
    var minimumValue = Infinity;
    var bestFactorCadidate = []
    for(let y=0; y<candidates.length; y++){
      var currentValue = 1;
      var currentPrime = 2;
      for(let z=0; z<combinedFactorCandidates[y].length; z++){
        currentValue *= Math.pow(currentPrime,(combinedFactorCandidates[y][z])-1);
        currentPrime = findPrimeAfter(currentPrime);
      }
      if(currentValue < minimumValue){
        minimumValue = currentValue;
        bestFactorCadidate = combinedFactorCandidates[y];
      }
    }
    return minimumValue;
  }

  var primeFactors = findPrimeFactors(factorCount);
  var primeFactorCombinations = removeDuplicateArrays(generateCombinations(primeFactors, 1));
  var combinedFactorCandidates = generateCombinedFactorCombinations(primeFactors, primeFactorCombinations);
  var smallestNumberWithFactorCount = determineMinimumCobination(combinedFactorCandidates);

  return smallestNumberWithFactorCount;
}

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Paste the above code block into your browser console and you can test it out yourself:

> smallestNumberByFactorCount(3) --> 4
> smallestNumberByFactorCount(4) --> 6
> smallestNumberByFactorCount(5) --> 16
> smallestNumberByFactorCount(6) --> 12
> smallestNumberByFactorCount(16) --> 120
> smallestNumberByFactorCount(100) --> 45360
> smallestNumberByFactorCount(500) --> 62370000
> smallestNumberByFactorCount(5000) --> 4727833110000
> smallestNumberByFactorCount(100000000) --> 1.795646397225103e+40

My algorithm starts shitting the bed when the input gets up to around 100 million... so for Project Euler problem 500 where the input would be 2^500500 (a really, really, really big number) you would need another approach. However, this is a good general approach that gets you pretty far. Hope it helps.

Please leave comments with efficiency optimization suggestions. I would love to hear them.