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R _ Q _ _ R _ D Clue
ROTATE BALL 9 TO MAKE INTO A 6. PLACE THE BALLS 6, 11 AND 13 TO GET A SUM OF 30. ALL THE NUMBERS ARE ODD. THERE IS NO WAY TO GET AN EVEN NUMBER, 30, BY SUMMING THREE ODD NUMBERS TOGETHER. AT LEAST ONE EVEN NUMBER IS REQUIRED THE NUMBERS CAN BE GROUPED BY PAIRS: 999,999,999 AND 0; 999,999,998 AND 1′ 999,999,997 AND 2; AND SO ON.... THERE ARE HALF A BILLION PAIRS, AND THE SUM OF THE DIGITS IN EACH PAIR IS 81. THE DIGITS IN THE UNPAIRED NUMBER, 1,000,000,000, ADD TO 1. THEN: (500,000,000 X 81) + 1= 40,500,000,001 THE PROFESSOR HAS TO ADD THE REST OF THE DIGITS, FIND THE NEAREST NUMBER TO THE SUM THAT IS DIVISIBLE BY 9 AND GET THE DIFFERENCE. SO, JOHN GAVE THE NUMBER 9646 TO THE PROFESSOR. THE PROFESSOR WILL ADD THE NUMBERS (9 + 6 + 4 + 6) TO GET 25. THE NEAREST NUMBER TO 25 THAT IS DIVISIBLE BY 9 IS 27. AND THE CROSSED OUT NUMBER IS 27 - 25. THIS IS A MATHS TRICK THAT RELIES ON THE POWER OF 9 ABEL HAS 50, BILL HAS 20 AND CLARK HAS 30. ABEL ON HIS FIRST TURN OBVIOUSLY DOESN'T KNOW WHETHER HIS NUMBER IS 50 OR 10. SIMILARLY NEITHER BILL NOR CLARK CAN IMMEDIATELY FIGURE OUT THEIR NUMBERS. HOWEVER, ON HIS SECOND TURN ABEL CAN REASON: IF MINE IS A 10, THEN CLARK WOULD KNOW HIS NUMBER IS EITHER 10 OR 30. IF IT IS 10, BILL WOULD IMMEDIATELY KNOW HIS NUMBER IS 20. BUT HE DIDN'T KNOW. SO CLARK SHOULD KNOW HIS NUMBER IS 30. NOW SINCE CLARK DIDN'T KNOW, MY NUMBER MUST BE 50. WITH THIS KIND OF REASONING WE CAN ALSO RULE OUT ALL OTHER COMBINATIONS. SO [50, 20, 30] IS THE ONLY SOLUTION TO THIS PUZZLE