Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• Karl - 10 years, 6 months ago

  Do you mean 210?

  3 and 7 aren't factors of 10 :P

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• Max Burstein - 10 years, 6 months ago

  Whoops I messed up on that one. I updated the problem

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• David - 10 years, 6 months ago

  I looked at the problem just in time to see your edits. Lucky me. ^{_^}

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• Anonymous - 10 years, 6 months ago

  #python 2.7
  def sum_prime_factors(n):
      i = 2
      prime_factors = []
      while i * i <= n:
          if n % i:
              i = i + 1
          else:
              n = n / i
              if i not in prime_factors:
                  prime_factors.append(i)
      if n > 1:
          if n not in prime_factors:
              prime_factors.append(n)    
      s = 0
  
      for factor in prime_factors:
          s += factor
  
      return s
  

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• Max Burstein - 10 years, 6 months ago

  Could always change prime_factors to a set to speed up look up time. Other then that it looks good. I like it.

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• Alex - 10 years, 6 months ago

  function primeSum(number){
      var sum = 0;
      for(var i = 2; i < number; i++){
          if(number % i == 0){
              sum += i;
          }
      }
      return sum;
  }
  

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• Mre64 - 10 years, 6 months ago

  perfectly simple, love it...

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• bumbleguppy - 10 years, 6 months ago

  javascript

    function sum_O_Primes(val)
    {
       var result = 0;
       for(var i = 2, mid = Math.ceil(val / 2); i <= mid; i++){
          if(val % i === 0 && isPrime(i)){
             result += i;
          }
       }  
  
       return result;
  
       function isPrime(num){
          if(num <= 1) return false;
          if(num === 2 || num === 3) return true;
          if(num % 2 === 0 || num % 3 === 0) return false;
  
          var squareRoot = Math.sqrt(num);
  
          for(var i = 3; i <= squareRoot; i+=2){
             if(num % i === 0) return false;
          }
          return true;
       }
    }
  
    console.log(sum_O_Primes(10));
  

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