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

Comments:

• parth - 10 years, 3 months ago

  54

  only one number.

  js:

  for (var i=10;i<10000000;i++){ if ( sumOfDigits(i)*6 == i ){ console.log(i); } }

  function sumOfDigits(d){ d = parseInt(d); var sum = 0; while (d > 0){ sum += d%10; d = parseInt(d/10); } return sum; }

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• Anonymous - 10 years, 3 months ago

  0 is the first, 54 is the second

  def sum_digits(n):
      return sum(map(int, str(n)))
  
  for i in range(1000000):
      if 6*sum_digits(i) == i:
          print(i)
  

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• Anonymous - 10 years, 3 months ago

  There are two such numbers : 0 and 54

  Negative and non-integer numbers are ruled out by the fact that the sum of the digits of a number is always a positive integer.

  So we are left with positive integers, for which we should solve the following indetermined sum (a[i] is the i-th digit of a (starting at i=0), and n is the number of digits of a):

  0 == \sum_{i = 0}^{n-1} (6 - 10ⁱ⁾ * a[i]

  Note that for i=0, (6 - 10ⁱ⁾ = 5, which is the only positive term.

  The next one is -4.

  Then, we are left with numbers < -90. Since 0 <= a[i] <= 9, those cannot be compensated by the first (positive) term (since 9*5 = 45).

  So the only possibilities are 0 and 54.

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• wobblebucket - 10 years, 3 months ago

  x = 1
  while x != (sum(map(int, str(x))) * 6):
      x = x + 1
  print x
  

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• btgrant-76 - 10 years, 3 months ago

  A little Scala. No state mutations and par produces a parallel collection.

    println((1 to 1000000).par
      .map { num: Int =>
        (num, num.toString.map {c: Char => c.asDigit }.sum)
      }.filter { numAndSum: (Int, Int) =>
        numAndSum._1 == numAndSum._2 * 6
      }.map { numAndSumTimesSix: (Int, Int) =>
        numAndSumTimesSix._1
      }.mkString(", "))
  

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• asheehan - 10 years, 2 months ago

  thought I submitted one for this, but don't see it here. Hmmm... Here's my program in Ruby:

  
  upperBound = ARGV[0].to_i
  for i in 0..upperBound
      digits = i.to_s.chars.map(&:to_i)
      sum = 0
      digits.each do |digit|
          sum += digit
      end
  
      if i == (sum * 6)
          puts i
      end
  end
  
  # output: 0, 54
  

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