Puzzle for August 26, 2023  ( )

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Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A + B + C + D + E + F = 30 eq.2) D = B + F eq.3) D = C + E eq.4) B + E = A + C eq.5) A + E = B + C + D + F

A, B, C, D, E, and F each represent a one-digit non-negative integer.

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Hint #1


eq.1 may be written as: B + F + C + E + A + D = 30 In the above equation, replace B + F with D (from eq.2), and C + E with D (from eq.3): D + D + A + D = 30 which becomes 3×D + A = 30 Subtract 3×D from both sides: 3×D + A - 3×D = 30 - 3×D which becomes eq.1a) A = 30 - 3×D


  

Hint #2


To make eq.1a true, check several possible values for D and A:   If D = 9, then A = 30 - 3×9 = 30 - 27 = 3 If D = 8, then A = 30 - 3×8 = 30 - 24 = 6 If D = 7, then A = 30 - 3×7 = 30 - 21 = 9 If D = 6, then A = 30 - 3×6 = 30 - 18 = 12 If D < 6, then A > 12   Since A must be a one-digit non-negative integer, then the above equations make: D ≥ 7 which may be written as ie.1b) 7 ≤ D


  

Hint #3


eq.1 may be written as: A + E + B + C + D + F = 30 In the equation above, replace A + E with B + C + D + F (from eq.5): B + C + D + F + B + C + D + F = 30 which becomes 2×(B + C + D + F) = 30 Divide both sides by 2: 2×(B + C + D + F) ÷ 2 = 30 ÷ 2 which becomes B + C + D + F = 15 which may be written as eq.1c) C + D + B + F = 15


  

Hint #4


In eq.1c, replace B + F with D (from eq.2): C + D + D = 15 which becomes C + 2×D = 15 Subtract 2×D from each side of the equation above: C + 2×D - 2×D = 15 - 2×D which becomes eq.1d) C = 15 - 2×D


  

Hint #5


To make eq.1d true, check several possible values for D and C:   If D = 9, then C = 15 - 2×9 = 15 - 18 = -3 If D = 8, then C = 15 - 2×8 = 15 - 16 = -1 If D = 7, then C = 15 - 2×7 = 15 - 14 = 1 If D < 7, then C > 1   Since C must be a one-digit non-negative integer, then the above equations make: ie.1e) D ≤ 7


  

Hint #6


Combining ie.1b with ie.1e yields: 7 ≤ D ≤ 7 which makes D = 7 and also makes A = 30 - 3×D = 30 - 3×7 = 30 - 21 = 9 (from eq.1a) C = 15 - 2×D = 15 - 2×7 = 15 - 14 = 1 (from eq.1d)


  

Hint #7


Substitute 7 for D, and 1 for C in eq.3: 7 = 1 + E Subtract 1 from both sides of the equation above: 7 - 1 = 1 + E - 1 which makes 6 = E


  

Hint #8


Substitute 6 for E, 9 for A, and 1 for C in eq.4: B + 6 = 9 + 1 which becomes B + 6 = 10 Subtract 6 from both sides of the above equation: B + 6 - 6 = 10 - 6 which makes B = 4


  

Solution

Substitute 7 for D, and 4 for B in eq.2: 7 = 4 + F Subtract 4 from each side of the above equation: 7 - 4 = 4 + F - 4 which makes 3 = F and makes ABCDEF = 941763