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A MUST NOT HAVE SEEN TWO WHITE HATS ON B AND C, OR HE WOULD HAVE KNOWN HIS OWN HAT MUST BE BLACK SINCE THERE ARE ONLY TWO WHITE HATS. SO A'S ANSWER ESTABLISHES THAT AT LEAST ONE OF B OR C'S HAT IS BLACK. BASED ON A'S ANSWER, B KNOWS THAT HE AND C ARE EITHER BOTH WEARING BLACK, OR ONE IS WEARING BLACK AND ONE IS WEARING A WHITE HAT. IF B SEES THAT C IS WEARING A WHITE HAT, THEN HE WOULD KNOW HIS OWN HAT HAD TO BE BLACK. BUT B DOES NOT KNOW WHAT COLOR HAT HE IS WEARING, WHICH MEAN'S C'S HAT IS NOT WHITE AND MUST BE BLACK. SINCE BOTH A AND B CANNOT DEDUCE THE COLOR OF THEIR OWN HATS, C WILL KNOW THAT HE IS WEARING A BLACK HAT 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 BILL IS THE QUARTERBACK, KEN IS THE RECEIVER, AND MARK IS THE KICKER. SINCE BILL IS KEN'S FATHER-IN-LAW, BOTH BILL AND KEN MUST BE MARRIED. THIS LEAVES THE ONLY REMAINING PERSON, MARK, TO BE THE BACHELOR (AND HENCE THE KICKER). SINCE BILL IS TALLER THAN THE RECEIVER, BILL MUST NOT BE THE RECEIVER. BY PROCESS OF ELIMINATION, WE CONCLUDE THAT BILL IS THE QUARTERBACK. THE REMAINING POSITION (THE RECEIVER) GOES TO KEN NO ONE IS TALLER, BOTH ARE SAME AS A AND B ARE THE SAME PERSON. AS IT IS MENTIONED THAT 500 MEN ARE ARRANGED IN AN ARRAY OF 10 ROWS AND 50 COLUMNS ACCORDING TO THEIR HEIGHTS. LET'S ASSUME THAT POSITION NUMBERS REPRESENT THEIR HEIGHTS. HENCE, THE SHORTEST AMONG THE 50, 100, 150, ... 450, 500 IS PERSON WITH HEIGHT 50 I.E. A. SIMILARLY THE TALLEST AMONG 1, 2, 3, 4, 5, ..... 48, 48, 50 IS PERSON WITH HEIGHT 50 I.E. B. NOW, BOTH A AND B ARE THE PERSON WITH HEIGHT 50. HENCE BOTH ARE THE SAME