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

Today's goal is to try and find an integer where the digit in the ones column is a 6 and when the 6 is moved to the front of the number, the number becomes 4 times the value of the starting number.

Comments:

• Anonymous - 9 years, 9 months ago

  153846

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• Jt - 9 years, 9 months ago

  bash 153,846

  #!/bin/bash
  counter=6
  
  while [ $counter -lt 1000000 ]; do
      if [[ ${counter:${#counter}-1:1} == '6' ]]
      then
        mult=$(($counter * 4))
        if [[ $mult == 6${counter:0:${#counter}-1} ]]
        then
          echo $counter $mult 6${counter:0:${#counter}-1}
          exit
        fi
      fi 
  
      let counter=counter+10
  done
  

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• Kocsen - 9 years, 9 months ago

  def mobile():
      LIMIT = 999999
      for number in range(6, LIMIT, 10):
          num_str = str(number)
          intified = int("6" + num_str[:-1])
          if intified == number * 4:
              print("Seems like we found it!")
              print(num_str)
              break
  
  
  
  if __name__ == "__main__":
      mobile()
  

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• Anonymous - 9 years, 9 months ago

  in java, i tried to calculate without using string

  static private void  runThis(){
  
      final int theMover = 6;
      final int theMultiplier = 4;
  
      for (int i = 1; i < 1000000; i++) {
          if (i % 10 == theMover) {
  
              // 6 * 10 power of x + remainder of 'i' 
              int j = theMover * (int) Math.pow(10, (int) Math.log10(i)) + (int) Math.floor(i / 10);
              if (i * theMultiplier == j) {
                  log.info("moving " + theMover + " across \t\t " + i + " * " + theMultiplier + " = " + j);
                  break;
              }
          }
      }
  }
  

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