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1 2 Clue
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 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 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 THE ADDRESS IS 1460 SUNSET BOULEVARD. YOU KNOW THAT THE HOUSE NUMBERS ARE EVEN AND CONSECUTIVE, SO THEY MUST BE APPROXIMATELY 1/6TH THE VALUE OF THE SUM 8790. IN FACT, THE NUMBER THAT IS 1/6TH THE TOTAL IS THE MEAN (AVERAGE) FOR ALL 6 HOUSES. THE AVERAGE NUMBER IS 1465 (8790 / 6). THERE MUST BE 3 HOUSE NUMBERS GREATER THAN THAT NUMBER, AND 3 HOUSE NUMBERS LESS THAN THAT NUMBER, ALL BEING EVEN AND CONSECUTIVE. THEREFORE, THE 6 HOUSE NUMBERS ARE 1460, 1462, 1464, 1466, 1468, 1470. THE LOWEST HOUSE NUMBER, AS PER THE QUESTION, IS THE ANSWER: 1460