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G _ N _ R _ L _ S _ T _ _ N
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K
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( _ N S T _ _ D
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≥
( 1
−
1 ⁄ P
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W H _ R _
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T H _
G R _ _ T _ S T
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N Clue
THE CHILDREN ARE 1, 6 AND 6 YEARS OLD. THE PRODUCT OF THEIR AGES IS 36, SO NONE OF THEM CAN BE OLDER THAN 36. THE NUMBER 36 HAS TO BE EXPRESSED AS THE PRODUCT OF 3 NUMBERS. THEIR POSSIBLE AGES ARE (THE SUM OF THEIR AGES IS IN BRACKETS): 1, 1, 36 (3938) 1, 2, 18 (21) 1, 3, 12 (16) 1, 4, 9 (14) 1, 6, 6 (13) 2, 2, 9 (13) 2, 3, 6 (11) 3, 3, 4 (10) SINCE CHERYL IS TOM'S NEXT DOOR NEIGHBOUR, TOM KNOWS CHERYL'S HOUSE NUMBER. TOM WOULD KNOW THE CHILDREN'S AGES IN EVERY CASE THAT SUMS UP TO A UNIQUE NUMBER EXCEPT FOR THE SUM OF 13, WHICH HAVE 2 COMBINATIONS OF POSSIBLE AGES. AS A RESULT, TOM WOULD BE CONFUSED AS HE HAS TO PICK BETWEEN THE 2 COMBINATIONS: (1,6,6) AND (2,2,9). CHERYL THEN TELLS TOM ABOUT HER YOUNGEST CHILD WHO LIKES STRAWBERRY MILK WHICH TELLS TOM THAT THERE IS ONLY 1 YOUNGEST CHILD BRUCE TAKES 9 OF HIS 10 CIGARETTE BUTTS AND TURNS THEM INTO 3 CIGARETTES TOTAL (3 CIGARETTE BUTTS CAN BE TURNED INTO 1 CIGARETTE). HE SMOKES ALL THREE OF THESE, AND NOW HE HAS 4 CIGARETTE BUTTS. HE THEN TURNS 3 OF THE 4 CIGARETTE BUTTS INTO ANOTHER CIGARETTE AND SMOKES IT. HE HAS NOW SMOKED 4 CIGARETTES AND HAS 2 CIGARETTE BUTTS. AND FINALLY HE GOES AND BORROWS ONE OF TOM'S CIGARETTE BUTTS. WITH THIS CIGARETTE BUTT PLUS THE 2 HE ALREADY HAS, HE IS ABLE TO MAKE HIS 5TH CIGARETTE TO SMOKE. AFTER SMOKING IT, HE IS LEFT WITH 1 CIGARETTE BUTT, WHICH HE PUTS BACK IN TOM'S PILE SO THAT TOM WON'T FIND ANYTHING MISSING WHEN THE FIRST SERVANT COMES IN, THE KING SHOULD WRITE DOWN HIS NUMBER. FOR EACH OTHER SERVANT THAT REPORTS IN, THE KING SHOULD ADD THAT SERVANT'S NUMBER TO THE CURRENT NUMBER WRITTEN ON THE PAPER, AND THEN WRITE THIS NEW NUMBER ON THE PAPER. LET X BE THE NUMBER OF THE MISSING SERVANT AND Y BE THE NUMBER THAT THE KING HAS WRITTEN. ONCE THE FINAL SERVANT HAS REPORTED IN, THE NUMBER ON THE PAPER SHOULD EQUAL: Y = (1 + 2 + 3 + ... + 99 + 100) - X (1 + 2 + 3 + ... + 99 + 100) = 5050, SO WE CAN REPHRASE THIS TO SAY THAT THE NUMBER ON THE PAPER SHOULD EQUAL: Y = 5050 - X SO TO FIGURE OUT THE MISSING SERVANT'S NUMBER, THE KING SIMPLY NEEDS TO SUBTRACT THE NUMBER WRITTEN ON HIS PAPER FROM 5050: 5050 - Y = X AN ALGORITHM THAT GUARANTEES THE BELL WILL RING IN AT MOST FIVE TURNS IS AS FOLLOWS: ON THE FIRST TURN CHOOSE A DIAGONALLY OPPOSITE PAIR OF GLASSES AND TURN BOTH GLASSES UP. ON THE SECOND TURN CHOOSE TWO ADJACENT GLASSES. AT LEAST ONE WILL BE UP AS A RESULT OF THE PREVIOUS STEP. IF THE OTHER IS DOWN, TURN IT UP AS WELL. IF THE BELL DOES NOT RING, THEN THERE ARE NOW THREE GLASSES UP AND ONE DOWN. ON THE THIRD TURN CHOOSE A DIAGONALLY OPPOSITE PAIR OF GLASSES. IF ONE IS DOWN, TURN IT UP AND THE BELL WILL RING. IF BOTH ARE UP, TURN ONE DOWN. THERE ARE NOW TWO GLASSES DOWN, AND THEY MUST BE ADJACENT. ON THE FOURTH TURN CHOOSE TWO ADJACENT GLASSES AND REVERSE BOTH. IF BOTH WERE IN THE SAME ORIENTATION THEN THE BELL WILL RING. OTHERWISE THERE ARE NOW TWO GLASSES DOWN AND THEY MUST BE DIAGONALLY OPPOSITE. ON THE FIFTH TURN CHOOSE A DIAGONALLY OPPOSITE PAIR OF GLASSES AND REVERSE BOTH. THE BELL WILL RING. THE PUZZLE CAN BE GENERALISED TO N GLASSES INSTEAD OF FOUR. FOR TWO GLASSES IT IS TRIVIALLY SOLVED IN ONE TURN BY INVERTING EITHER GLASS. FOR THREE GLASSES THERE IS A TWO-TURN ALGORITHM. FOR FIVE OR MORE GLASSES THERE IS NO ALGORITHM THAT GUARANTEES THE BELL WILL RING IN A FINITE NUMBER OF TURNS. A FURTHER GENERALISATION ALLOWS K GLASSES (INSTEAD OF TWO) OUT OF THE N GLASSES TO BE EXAMINED AT EACH TURN. AN ALGORITHM CAN BE FOUND TO RING THE BELL IN A FINITE NUMBER OF TURNS AS LONG AS K ≥ (1 − 1⁄P )N WHERE P IS THE GREATEST PRIME FACTOR OF N