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1 2 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 FIRST HEAP: 22, SECOND HEAP: 14, THIRD HEAP: 12. TO SOLVE THIS PROBLEM, WE SHALL HAVE TO START FROM THE END. WE HAVE BEEN TOLD THAT AFTER ALL THE TRANSPOSITIONS, THE NUMBER OF MATCHES IN EACH HEAP IS THE SAME. LET US PROCEED FROM THIS FACT. SINCE THE TOTAL NUMBER OF MATCHES HAS NOT CHANGED IN THE PROCESS, AND THE TOTAL NUMBER BEING 48, IT FOLLOWS THAT THERE WERE 16 MATCHES IN EACH HEAP. AND SO, IN THE END WE HAVE: FIRST HEAP: 16, SECOND HEAP: 16, THIRD HEAP: 16 IMMEDIATELY BEFORE THIS WE HAVE ADDED TO THE FIRST HEAP AS MANY MATCHES AS THERE WERE IN IT, I.E. WE HAD DOUBLED THE NUMBER. SO, BEFORE THE FINAL TRANSPOSITION, THERE ARE ONLY 8 MATCHES IN THE FIRST HEAP. NOW, IN THE THIRD HEAP, FROM WHICH WE TOOK THESE 8 MATCHES, THERE WERE: 16 + 8 = 24 MATCHES. WE NOW HAVE THE NUMBERS AS FOLLOWS: FIRST HEAP: 8, SECOND HEAP: 16, THIRD HEAP: 24. WE KNOW THAT WE TOOK FROM THE SECOND HEAP AS MANY MATCHES AS THERE WERE IN THE THIRD HEAP, WHICH MEANS 24 WAS DOUBLE THE ORIGINAL NUMBER. FROM THIS WE KNOW HOW MANY MATCHES WE HAD IN EACH HEAP AFTER THE FIRST TRANSPOSITION: FIRST HEAP: 8, SECOND HEAP: 16 + 12 = 28, THIRD HEAP: 12. NOW WE CAN DRAW THE FINAL CONCLUSION THAT BEFORE THE FIRST TRANSPOSITION THE NUMBER OF MATCHES IN EACH HEAP WAS: FIRST HEAP: 22, SECOND HEAP: 14, THIRD HEAP: 12 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 YOU WOULD ONLY NEED TO TAKE OUT ONE MARBLE BECAUSE WE KNOW THAT ALL OF THE LABELS ARE INCORRECT. SO YOU PULL ONE MARBLE OUT OF THE BOX LABELED "MIXED." IF RED COMES OUT, YOU KNOW THAT HAS TO BE THE ALL-RED BOX, SO YOU PUT THE RED LABEL ON IT. THE BOX LABELLED "BLUE" MUST THEN BE LABELLED "MIXED" BECAUSE YOU KNOW IT IS ALSO LABELED INCORRECTLY, AND THEREFORE CAN'T BE BLUE. YOU WOULD LABEL THE LAST BOX "BLUE" BECAUSE THAT IS THE ONLY COLOR/BOX COMBO LEFT. IF THE FIRST MARBLE YOU PULLED OUT FROM THE BOX LABELED "MIXED" IS A BLUE MARBLE, THEN YOU SOLVE THE PROBLEM IN THE SAME GENERAL WAY