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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 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 MR TAN'S BIRTHDAY IS ON 1/9/1970. N AND M REPRESENT THE DAY AND MONTH OF A DATE RESPECTIVELY. BEN CAN ONLY KNOW THE DATE IF HE IS GIVEN A UNIQUE M. THERE ARE NO UNIQUE M'S. MARK CAN ONLY KNOW THE DATE IF HE IS GIVEN A UNIQUE N. THERE ARE TWO UNIQUE N'S: 7 AND 2. SINCE BEN CAN ENSURE THAT MARK DOESN'T KNOW, WE KNOW MARK CAN'T HAVE ANY M THAT CORRESPONDS TO THE UNIQUE N'S. SO, THIS ELIMINATES JUNE AND DECEMBER. WE ARE NOW LEFT WITH MARCH AND SEPTEMBER: 4/3/1970 5/3/1970 8/3/1970 1/9/1970 5/9/1970 SO MARK EITHER HAS 1, 4, 5, OR 8. SINCE MARK IS ABLE TO DETERMINE THE DATE, THE DAY HAS TO BE UNIQUE. IF MARK WAS TOLD 1, HE KNOWS IT'S 1/9/1970. IF HE WAS TOLD 4, IT'S 4/3/1970. IF HE GOT 8, IT'S 8/3/1970. IF HE HAD BEEN TOLD 5, THERE IS NOT ENOUGH INFORMATION AS IT IS NOT A UNIQUE, SO IT CANNOT BE 5. SO, WE ARE NOW LEFT WITH: 4/3/1970 8/3/1970 1/9/1970 AND FINALLY, SINCE BEN CAN IDENTIFY THE BIRTH DATE AT THIS POINT, THIS MEANS THAT THERE MUST BE ONLY ONE UNIQUE MONTH VALUE LEFT WHICH WOULD ELIMINATE 4/3/1970 AND 8/3/1970, LEAVING 1/9/1970 AS THE CORRECT BIRTH DATE 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