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_ R R _ R S Clue
IT IS OBVIOUS THAT THERE MUST BE AN ODD NUMBER OF CROSSINGS, AND THAT IF THE FIVE HUSBANDS HAD NOT BEEN JEALOUS OF ONE ANOTHER THE PARTY MIGHT HAVE ALL GOT OVER IN NINE CROSSINGS. BUT NO WIFE WAS TO BE IN THE COMPANY OF A MAN OR MEN UNLESS HER HUSBAND WAS PRESENT. THIS ENTAILS TWO MORE CROSSINGS, ELEVEN IN ALL. THE FOLLOWING SHOWS HOW IT MIGHT HAVE BEEN DONE. THE CAPITAL LETTERS STAND FOR THE HUSBANDS, AND THE SMALL LETTERS FOR THEIR RESPECTIVE WIVES. THE POSITION OF AFFAIRS IS SHOWN AT THE START, AND AFTER EACH CROSSING BETWEEN THE LEFT BANK AND THE RIGHT, AND THE BOAT IS REPRESENTED BY THE ASTERISK. SO YOU CAN SEE AT A GLANCE THAT A, B, AND C WENT OVER AT THE FIRST CROSSING, THAT B AND C RETURNED AT THE SECOND CROSSING, AND SO ON. ABCDE ABCDE *- ABCDE DE-* ABC ABCDE BCDE *-A ABCDE E-* ABCDE ABCDE DE *-ABC DE DE-* ABC ABC CDE CDE *-AB AB CDE-* ABCDE AB BCDE *-ABCDE A E-* ABCDE ABCDE BC E *-ABCDE AD -* ABCDE ABCDE THERE IS A LITTLE SUBTLETY CONCEALED IN THE WORDS "SHOW THE QUICKEST WAY." EVERYBODY CORRECTLY ASSUMES THAT, AS WE ARE TOLD NOTHING OF THE ROWING CAPABILITIES OF THE PARTY, WE MUST TAKE IT THAT THEY ALL ROW EQUALLY WELL. BUT IT IS OBVIOUS THAT TWO SUCH PERSONS SHOULD ROW MORE QUICKLY THAN ONE. THEREFORE IN THE SECOND AND THIRD CROSSINGS TWO OF THE LADIES SHOULD TAKE BACK THE BOAT TO FETCH D, NOT ONE OF THEM ONLY. THIS DOES NOT AFFECT THE NUMBER OF LANDINGS, SO NO TIME IS LOST ON THAT ACCOUNT. A SIMILAR OPPORTUNITY OCCURS IN CROSSINGS 10 AND 11, WHERE THE PARTY AGAIN HAD THE OPTION OF SENDING OVER TWO LADIES OR ONE ONLY. TO THOSE WHO THINK THEY HAVE SOLVED THE PUZZLE IN NINE CROSSINGS I WOULD SAY THAT IN EVERY CASE THEY WILL FIND THAT THEY ARE WRONG. NO SUCH JEALOUS HUSBAND WOULD, IN THE CIRCUMSTANCES, SEND HIS WIFE OVER TO THE OTHER BANK TO A MAN OR MEN, EVEN IF SHE ASSURED HIM THAT SHE WAS COMING BACK NEXT TIME IN THE BOAT. IF READERS WILL HAVE THIS FACT IN MIND, THEY WILL AT ONCE DISCOVER THEIR ERRORS LET'S ASSUME THAT THERE IS ONLY 1 CHEATING HUSBAND. THEN HIS WIFE DOESN'T SEE ANYBODY CHEATING, SO SHE KNOWS HE CHEATS, AND SHE WILL KILL HIM THAT VERY DAY. NOW, LET'S SAY THAT THERE ARE 2 CHEATING HUSBANDS. THERE WILL BE 98 WOMEN IN THE TOWN WHO KNOW WHO THE 2 CHEATERS ARE. THE 2 WIVES, WHO ARE BEING CHEATED ON, WOULD THINK THAT THERE IS ONLY 1 CHEATER IN THE TOWN. SINCE NEITHER OF THESE 2 WOMEN KNOW THAT THEIR HUSBANDS ARE CHEATERS, THEY BOTH DO NOT REPORT THEIR HUSBANDS IN ON THE DAY OF THE ANNOUNCEMENT. THE NEXT DAY, WHEN THE 2 WOMEN SEE THAT NO HUSBAND WAS EXECUTED, THEY REALIZE THAT THERE COULD ONLY BE ONE EXPLANATION - BOTH THEIR HUSBANDS ARE CHEATERS. THUS, ON THE SECOND DAY, 2 HUSBANDS ARE EXECUTED. THROUGH MATHEMATICAL INDUCTION, IT CAN BE PROVED THAT WHEN THIS LOGIC IS APPLIED TO N CHEATING HUSBANDS, THEY ALL DIE ON THE N TH DAY AFTER THE QUEEN'S ANNOUNCEMENT. SO WITH 100 CHEATING HUSBANDS, ALL OF THEM WILL BE EXECUTED ON THE 100TH DAY THE ELDEST IS 8 YEARS OLD AND THE 2 YOUNGER ONES ARE 3 YEARS OLD. LET'S BREAK IT DOWN. THE PRODUCT OF THEIR AGES IS 72. SO THE POSSIBLE CHOICES ARE: 2, 2, 18 - SUM(2, 2, 18) = 22 2, 4, 9 - SUM(2, 4, 9) = 15 2, 6, 6 - SUM(2, 6, 6) = 14 2, 3, 12 - SUM(2, 3, 12) = 17 3, 4, 6 - SUM(3, 4, 6) = 13 3, 3, 8 - SUM(3, 3, 8 ) = 14 1, 8, 9 - SUM(1,8,9) = 18 1, 3, 24 - SUM(1, 3, 24) = 28 1, 4, 18 - SUM(1, 4, 18) = 23 1, 2, 36 - SUM(1, 2, 36) = 39 1, 6, 12 - SUM(1, 6, 12) = 19 THE SUM OF THEIR AGES IS THE SAME AS YOUR BIRTH DATE. THAT COULD BE ANYTHING FROM 1 TO 31 BUT THE FACT THAT JACK WAS UNABLE TO FIND OUT THE AGES, IT MEANS THERE ARE TWO OR MORE COMBINATIONS WITH THE SAME SUM. FROM THE CHOICES ABOVE, ONLY TWO OF THEM ARE POSSIBLE NOW. 2, 6, 6 - SUM(2, 6, 6) = 14 3, 3, 8 - SUM(3, 3, 8 ) = 14 SINCE THE ELDEST KID IS TAKING PIANO LESSONS, WE CAN ELIMINATE COMBINATION 1 SINCE THERE ARE TWO ELDEST ONES. THE ANSWER IS 3, 3 AND 8 THEIR HATS ARE ALL BLUE IN COLOR. THERE ARE 3 POSSIBLE HAT COLOR COMBINATIONS: [A] 1 BLUE, 2 WHITE [B] 2 BLUE, 1 WHITE [C] 3 BLUE THE COLOR COMBINATION OF 3 WHITE HATS IS NOT POSSIBLE SINCE THE KING HAS ALREADY SAID THAT AT LEAST ONE OF THE WISE MEN HAS A BLUE HAT. SO, LET'S START OUR ANALYSIS. WHAT IF THERE WERE ONE BLUE HAT AND TWO WHITE HATS? THEN THE WISE MAN WITH THE BLUE HAT WOULD HAVE SEEN TWO WHITE HATS AND IMMEDIATELY CALLED OUT THAT HIS OWN HAT WAS BLUE, SINCE HE KNEW THERE IS AT LEAST ONE BLUE HAT. THIS DIDN'T HAPPEN, AND THUS THE HAT COLOR COMBINATION [A] IS RULED OUT. NOW, THE WISE MEN KNEW THAT ONLY 2 HAT COLOR COMBINATIONS ARE POSSIBLE - COMBINATION [B] OR [C]. WHAT IF THERE WERE TWO BLUE HATS AND ONE WHITE HAT? THE MEN WITH THE BLUE HATS WILL SEE ONE WHITE HAT AND ONE BLUE HAT. THEY WILL CONCLUDE THAT COLOR COMBINATION [B] IS THE CASE AND WOULD CALL OUT BLUE AS THEIR HAT COLOR. THIS ALSO DID NOT HAPPEN, AND THUS THE HAT COLOR COMBINATION [B] IS ALSO RULED OUT. AFTER SOME TIME, WHEN NONE OF THE WISE MEN ARE ABLE TO IDENTIFY THE COLOR OF THEIR OWN HATS, COMBINATION [C] (OF 3 BLUE HATS) BECOMES THE ONLY POSSIBLE OPTION