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
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public class Main
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{
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public static void main(String[] args)
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{
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for(int i = 100; i <= 999; i++)
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{
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//get the number, and "double" it
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int number = Integer.parseInt(Integer.toString(i) + Integer.toString(i));
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//successively divided by 7, 11, then 13
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System.out.print((i) + "\t");
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System.out.print((number) + "\t");
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System.out.print((number /= 7) + "\t");
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System.out.print((number /= 11) + "\t");
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System.out.print(number /= 13);
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//is the result what we expect? (input == output)
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if(i == number)
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{
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System.out.print("\tCorrect\n");
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}
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else
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{
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System.out.print("\tIncorrect");
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break;
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}
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}
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}
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}
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!
