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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 MESSENGER HAS TO HAVE TRAVELED 2 KM. IT DOESN'T MATTER WHAT SPEED THEY WALKED AT. AT THE BEGINNING OF THE PUZZLE, THE LINE IS 1 KM LONG. THE GENERAL IS THEREFORE 1 KM AHEAD OF HIM. THE MESSENGER MUST THEREFORE TRAVEL MORE THAN 1 KM TO REACH THE GENERAL. SINCE THE LINE MOVES 1 KM FORWARD, THE END IS WHERE THE BEGINNING WAS. EVEN IF HE WALKED 1.5 KM TO THE GENERAL, HE ONLY HAS TO WALK 0.5 KM TO GET BACK TO THE END OF THE LINE. IT GOES FASTER GOING BACK, BECAUSE NOW THEY ARE COMING TOWARDS HIM, AND NOT GOING AWAY THEY STOLE 301 DIAMONDS IN TOTAL. WE NEED A NUMBER THAT IS A MULTIPLE OF 7 THAT WILL GIVE A REMAINDER OF 1 WHEN DIVIDED BY 2, 3, 4, 5, AND 6. THE LEAST COMMON MULTIPLE OF THESE NUMBERS IS 60. SO, WE NEED A MULTIPLE OF 7 THAT IS 1 GREATER THAN A MULTIPLE OF 60. 60 + 1 = 61, NOT A MULTIPLE OF 7 60 X 2 + 1 = 121, NOT A MULTIPLE OF 7 60 X 3 + 1 = 181, NOT A MULTIPLE OF 7 60 X 4 + 1 = 241, NOT A MULTIPLE OF 7 60 X 5 + 1 = 301, A MULTIPLE OF 7 TOM WINS AGAIN. IN THE FIRST RACE TOM RAN 100 METERS IN THE TIME IT TOOK HARRY TO RUN 95 METERS. SO IN THE SECOND RACE WHEN HARRY IS AT THE 95 METER MARK TOM WILL ALSO BE THERE (SINCE 100 - 5 = 95). SINCE TOM IS FASTER HE WILL PASS HARRY IN THE LAST 5 METERS OF THE RACE