“Presto” math trick in Java

According to Futility Closet, if you start with a 3 digit and place it next to the same number to form a 6 digit number, you can divide the 6 digit number by 7, 11, and then 13 and you will end up with the original 3 digit number and no remainders. For example, by taking the number 412 and making it 412412 and then doing the divisions, you will end up with 412. I wrote a small program in Java to test it.

The Code

public class Main
{
  public static void main(String[] args)
  {
    for(int i = 100; i <= 999; i++)
    {
      //get the number, and "double" it
      int number = Integer.parseInt(Integer.toString(i) + Integer.toString(i));

      //successively divided by 7, 11, then 13
      System.out.print((i) + "\t");
      System.out.print((number) + "\t");
      System.out.print((number /= 7) + "\t");
      System.out.print((number /= 11) + "\t");
      System.out.print(number /= 13);

      //is the result what we expect? (input == output)
      if(i == number)
      {
        System.out.print("\tCorrect\n");
      }
      else
      {
        System.out.print("\tIncorrect");
        break;
      }
    }
  }
}

In this snippet, we loop through all 3 digit numbers (100 to 999) and output the results of each operation in a tab separated column. If the result (output) is the same as the input, we print correct and continue with the loop. Otherwise, we print incorrect and break. Give it a try!

  1. Daniel Fitzpatrick

    (filter #(not= %1 (/ (/ (/ (* 1000 %1) 7) 11) 13)) (range 100 1000))

    ^untested, should do the same thing in clojure.

  2. Daniel Fitzpatrick

    Whoops. Got my logic backward. Switch the not= with = or filter with remove.

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