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S T _ R T Clue
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 25 EGGS WERE BROKEN. THERE IS ONLY ONE WAY OF FINDING A SOLUTION TO THIS PROBLEM: NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 2: 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35 NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 3: 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37 NUMBERS WHICH LEAVE A REMAINDER OF 1, WHEN DIVIDED BY 4: 5, 9, 13, 17, 21, 25, 29, 33, 37 NUMBERS WHICH LEAVE NO REMAINDER WHEN DIVIDED BY 5: 5, 10, 15, 20, 25, 30, 35 THE ONLY NUMBER FULFILLING THE FOUR CONDITIONS IS 25 SABRINA HAD $50 AND SAMANTHA HAD $30. THIS CAN BE SOLVED BY SETTING UP TWO SIMULTANEOUS EQUATIONS, BUT IT'S EASIER JUST TO WORK BACKWARDS. AT EACH STEP, THE PERSON WHO IS RECEIVING THE MONEY GETS THE AMOUNT THAT THEY WERE ALREADY HOLDING. IN OTHER WORDS, THEY DOUBLE THEIR MONEY. THEREFORE, JUST BEFORE THE LAST EXCHANGE, SAMANTHA MUST HAVE HAD $40 AND WAS GIVEN ANOTHER $40 TO GET TO HER TOTAL OF $80. JUST BEFORE THE SECOND EXCHANGE, SABRINA MUST HAVE HAD HALF OF HER $40. THEREFORE, SABRINA HAD $20 AND SAMANTHA HAD $60. AFTER THE FIRST EXCHANGE SAMANTHA DOUBLED HER MONEY, SO SHE MUST HAVE HAD $30 BEFORE THE EXCHANGE, LEAVING SABRINA WITH $50 AT THE START THE CONTENTS OR THE TEN ENVELOPES (IN DOLLAR BILLS) SHOULD BE AS FOLLOWS: $1, 2, 4, 8, 16, 32, 64, 128, 256, 489. THE FIRST NINE NUMBERS ARE IN GEOMETRICAL PROGRESSION, AND THEIR SUM, DEDUCTED FROM 1,000, GIVES THE CONTENTS OF THE TENTH ENVELOPE