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Nightlong Hints
How to solve the equation
1 of 10: The laboratory door access code: (1) you know that this code is 446677 (2) the symbols for this code don't appear on the blackboard (3) the numbers 4, 6 and 7 are excluded from the equation
2 of 10: The equation can be divided into two parts: PART ONE: abc x dc = defc PART TWO: defc + cbfc = agecb
3 of 10: From PART TWO: defc + cbfc = agecb "agecb" is a number in the "ten thousands" "agecb" cannot be 20 000 or greater "agecb" must be a number between 10 000 and 20 000 therefore a=1
4 of 10: From PART ONE: abc x dc = defc "c"' is the last digit in all the numbers so: c x c = c (or a number ending in "c") therefore "c" can only be 0,1,5 or 6 but "c" cannot be 1 (a=1) nor 6 (excluded) therefore "c" can only be 0 or 5
5 of 10: Back to PART TWO: defc + cbfc = agecb again look at the last digit in the numbers so: c + c = b (or a number ending in "b") but "c" cannot be 0 (0 + 0 = 0 doesn't fit) therefore c=5 also if: 5 + 5 = 10 (c + c = b) then b=0
6 of 10: Staying with PART TWO: defc + cbfc = agecb (def5 + 50f5 = 1ge50) look at the last two digits in the numbers so: f5 + f5 = 50 (or a number ending in "50") therefore "f" can only be 2 or 7 but "f" cannot be 7 (excluded) therefore f=2
7 of 10: What do you know so far? a=1 b=0 c=5 f=2 missing letters are "d", "e" and "g" missing numbers are 3, 8 and 9
8 of 10: From PART ONE: abc x dc = defc (105 x d5 = de25) and using "trial and error": if "d" is 3: 105 x 35 = 3e25 but 105 x 35 = 3675, so "d" cannot be 3 if "d" is 9: 105 x 95 = 9e25 but 105 x 95 = 9975, so "d" cannot be 9 if "d" is 8: 105 x 85 = 8e25 but 105 x 85 = 8925, so "d" can only be 8 therefore d=8 and e=9 also by process of elimination g=3
9 of 10: The governor's door access code from the piece of paper from Eva is "e c d g f b"
10 of 10: Therefore the code is 9 5 8 3 2 0. Congratulations!
