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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 PROBLEM IS THAT THE QUESTION IS CLEVERLY PHRASED TO CONCEAL WHAT IS REALLY GOING ON. LET'S LOCATE ALL THAT MONEY. THERE ARE TWO WAYS TO THINK ABOUT HOW MUCH MONEY IS OUT THERE TO BE FOUND.THE WAY THAT THIS QUESTION IS TRICKY IS THAT IT COMBINES THAT TWO WAYS. THE FIRST WAY IS THIS. HOW MUCH MONEY DID THE THREE MEN PAY ORIGINALLY? THE SECOND WAY IS THIS. HOW MUCH MONEY DID THEY END UP PAYING? SO IF IT IS THE FIRST WAY, THEN CLEARLY THE TOTAL WE NEED TO ACCOUNT FOR IS THE $15. SO LET'S SEE WHAT HAPPENS TO THAT $15. THE CHEF GETS $10, THE WAITER GETS $2 AND THE GUYS GET $3 BACK. THAT ADDS UP FINE. NOW LET'S LOOK AT THE SECOND WAY. HOW MUCH MONEY DID THEY END UP PAYING? WELL $12, AND $10 OF IT WENT TO THE CHEF AND $2 TO THE WAITER. NOW, THAT ADDS UP TOO. THE PROBLEM WITH THE QUESTION IS THAT THE THE $2 THAT THE WAITER TOOK IS CONTAINED IN THE $12 THAT THEY END UP PAYING SO WE SHOULDN'T EXPECT THEM TO ADD TO ANYTHING MEANINGFUL THE PROBLEM IS THAT THE QUESTION IS CLEVERLY PHRASED TO CONCEAL WHAT IS REALLY GOING ON. LET'S LOCATE ALL THAT MONEY. THERE ARE TWO WAYS TO THINK ABOUT HOW MUCH MONEY IS OUT THERE TO BE FOUND.THE WAY THAT THIS QUESTION IS TRICKY IS THAT IT COMBINES THAT TWO WAYS. THE FIRST WAY IS THIS. HOW MUCH MONEY DID THE GUYS PAY ORIGINALLY? THE SECOND WAY IS THIS. HOW MUCH MONEY DID THEY END UP PAYING? SO IF IT IS THE FIRST WAY, THEN CLEARLY THE TOTAL WE NEED TO ACCOUNT FOR IS THE $30. SO LET'S SEE WHAT HAPPENS TO THAT $30. THE MANAGER KEEPS $25, THE BELLBOY GETS $2 AND THE GUYS GET $3 BACK. THAT ADDS UP FINE. NOW LET'S LOOK AT THE SECOND WAY. HOW MUCH MONEY DID THEY END UP PAYING? WELL $27 DOLLARS, AND $25 OF IT WENT TO THE OWNER AND $2 TO THE BELLBOY. NOW, THAT ADDS UP TOO. THE PROBLEM WITH THE QUESTION IS THAT THE THE $2 THAT THE BELLBOY GETS ARE CONTAINED IN THE $27 THAT THEY END UP PAYING SO WE SHOULDN'T EXPECT THEM TO ADD TO ANYTHING MEANINGFUL 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