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1 7 Clue
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 WEIGHT OF THE 5 RINGS ARE 1, 2, 4, 8 AND 16 GRAMS. USING THE COMBINATION OF THE 5 TYPE OF WEIGHTS YOU CAN REWARD FROM 1 TO 31 GRAMS IN WEIGHT TO THE WISE MAN. FOR EXAMPLE: FOR THE 3RD DAY, YOU CAN GIVE HIM THE 1 AND 2 GRAM RINGS. FOR THE 15TH DAY, YOU CAN GIVE HIM THE 1, 2, 4, 8 GRAM RINGS. FOR THE 30TH DAY, YOU CAN GIVE HIM THE 2, 4, 8, 16 GRAMS 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 CHERYL'S BIRTHDAY IS ON JULY 16. FROM ALBERT'S FIRST STATEMENT, WE CAN ELIMINATE MAY AND JUNE. THE DATES MAY 19 AND JUNE 18 ARE UNIQUE (19 ONLY APPEARS IN MAY 19 AND 18 ONLY APPEARS IN JUNE 18) AND IF THERE IS NO WAY FOR BERNARD TO KNOW CHERYL'S BIRTHDAY, THEN IT CANNOT BE MAY OR JUNE. BERNARD CAN NOW ELIMINATE MAY AND JUNE AFTER ALBERT'S FIRST STATEMENT. IF BERNARD CAN NOW DEDUCE CHERYL'S BIRTHDAY, THE DATE HAS TO BE UNIQUE. THEREFORE, WE CAN ELIMINATE JULY 14 AND AUG 14 AS WELL. NOW WE ARE LEFT WITH JULY 16, AUG 15 OR AUG 17. ALBERT NOW KNOWS IT CAN ONLY BE JULY 16, AUG 15 OR AUG 17. HE KNOWS THE MONTH BUT NOT THE DATE, AND THAT IS ENOUGH TO TELL HIM THE CORRECT ANSWER. THEREFORE IT CAN ONLY BE JULY 16. IF IT HAD BEEN AUGUST, THEN HE WILL NOT KNOW IF IT IS AUG 15 OR AUG 17