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6 4 = 6 + 4 = 1 0 ,
8 4 = 8 + 4 = 1 2 ,
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8 6 = 8 + 6 = 1 4 .
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8 4 Clue
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 IF YOU FOLLOWED ALL THE STEPS APART FROM THE LAST ONE THERE WILL BE THREE OPTIONS REMAINING: 64, 84, AND 86. YOU THEN HAD TO ADD UP THE DIGITS, 64=6+4=10, 84=8+4=12, AND 86=8+6=14. FINALLY, YOU THEN HAD TO TAKE UP THE MIDDLE BIGGEST NUMBER (12) AND PUT IT BACK AS IT WAS BEFORE THE DIGITS WERE ADDED TOGETHER AND YOUR ANSWER SHOULD BE 84 THE SECRET CODE IS 7,4,6,5,8. LET THE NUMBERS = A, B, C, D, E BASED ON 1ST CLUE: E + C = 14 BASED ON 2ND CLUE: A = 2B - 1 BASED ON 3RD CLUE: D = B + 1 BASED ON 4TH CLUE: B + C = 10 BASED ON 5TH CLUE: A + B + C + D + E = 30 ALL NUMBERS CAN BE REPRESENTED BY B EXCEPT FOR E SO WE NEED TO REWRITE EQUATION 1. SUBSTITUTE C = 10 - B FROM EQUATION 4 INTO EQUATION 1. E + (10 - B) = 14 E = B + 4 SUBSTITUTE A,C,D,E IN EQUATION 5 (2B - 1) + B + (10 - B) + (B + 1) + (B + 4) = 30 SOLVING FOR B GIVES 4. USE THE OTHER EQUATIONS TO FIND THE VALUES OF A,C,D AND E THE NUMBERS CAN BE GROUPED BY PAIRS: 999,999,999 AND 0; 999,999,998 AND 1′ 999,999,997 AND 2; AND SO ON.... THERE ARE HALF A BILLION PAIRS, AND THE SUM OF THE DIGITS IN EACH PAIR IS 81. THE DIGITS IN THE UNPAIRED NUMBER, 1,000,000,000, ADD TO 1. THEN: (500,000,000 X 81) + 1= 40,500,000,001