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1 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 THE STUDENT IS DOUBLE COUNTING A LOT OF THE DAYS. A LOT OF THE TIME SPENT SLEEPING, EATING, AND RELAXING OCCURS DURING WEEKENDS AND THE SUMMER. WEEKENDS ALSO OCCUR DURING THE SUMMER, SO ALL OF THESE HOURS ARE GETTING COUNTED SEVERAL TIMES. AND, SCHOOL IS NOT AN ALL DAY AFFAIR. SO THE 4 DAYS ACTUALLY REPRESENTS MORE DAYS OF SCHOOL. IF SCHOOL IS 6 HOURS PER DAY, THOSE FOUR DAYS REPRESENTS 16 DAYS OF SCHOOL THE PROFESSOR HAS TO ADD THE REST OF THE DIGITS, FIND THE NEAREST NUMBER TO THE SUM THAT IS DIVISIBLE BY 9 AND GET THE DIFFERENCE. SO, JOHN GAVE THE NUMBER 9646 TO THE PROFESSOR. THE PROFESSOR WILL ADD THE NUMBERS (9 + 6 + 4 + 6) TO GET 25. THE NEAREST NUMBER TO 25 THAT IS DIVISIBLE BY 9 IS 27. AND THE CROSSED OUT NUMBER IS 27 - 25. THIS IS A MATHS TRICK THAT RELIES ON THE POWER OF 9 THE TOTAL IS 2. WHEN THE BOOK ON THE LEFT IS UPSIDE DOWN THEN THE PAGE NUMBER ON THE EXTREME LEFT IS 1 AND THE RIGHT HAND SIDE BOOK'S EXTREME RIGHT HAND SIDE PAGE NUMBER IS ALSO 1. ANOTHER WAY TO LOOK AT THE PROBLEM: IF TWO BOOKS ARE STANDING SIDE BY SIDE TO EACH OTHER, PAGE 1 IS ON THE RIGHT SIDE OF EACH BOOK (ASSUMING YOU ARE LOOKING AT THE SPINE OF THE BOOKS). TURNING THE ONE ON THE LEFT UPSIDE DOWN (WITH THE BOOK SPINE STILL POINTING OUT), BRINGS PAGE 1 TO THE LEFT SIDE. AND SO AT THE "EXTREME LEFT" OF THE BOOK ON THE LEFT IS PAGE 1, WHILE THE "EXTREME RIGHT" OF THE BOOK ON THE RIGHT IS ALSO PAGE 1