Puzzle for August 13, 2020  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) B + F = A + D eq.3) C + E = A – D eq.4) E – F = B – C – D eq.5) B + C + F = A – C + E eq.6) E + F – B = A + B

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

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Hint #1


Add D to both sides of eq.3: C + E + D = A – D + D which becomes C + E + D = A which may be written as eq.3a) C + D + E = A   Add C, D, and F to each side of eq.4: E – F + C + D + F = B – C – D + C + D + F which becomes E + C + D = B + F which may be written as eq.4a) C + D + E = B + F


  

Hint #2


In eq.3a, replace C + D + E with B + F (from eq.4a): eq.3b) B + F = A   In eq.2, replace B + F with A (from eq.3b): A = A + D Subtract A from both sides of the equation above: A – A = A + D – A which means 0 = D


  

Hint #3


In eq.3a, substitute 0 for D: C + 0 + E = A which becomes eq.3c) C + E = A


  

Hint #4


eq.5 may be written as: B + F + C = A – C + E In the above equation, substitute A for B + F (from eq.3b): A + C = A – C + E Subtract A from both sides, and add C to both sides: A + C – A + C = A – C + E – A + C which makes 2×C = E


  

Hint #5


Substitute 2×C for E in eq.3c: C + 2×C = A which makes 3×C = A


  

Hint #6


Substitute 3×C for A in eq.3b: B + F = 3×C Subtract B from each side of the equation above: B + F – B = 3×C – B which becomes eq.3d) F = 3×C – B


  

Hint #7


Substitute 2×C for E, 3×C – B for F (from eq.3d), and 3×C for A in eq.6: 2×C + 3×C – B – B = 3×C + B which becomes 5×C – 2×B = 3×C + B In the above equation, add 2×B to each side, and subtract 3×C from each side: 5×C – 2×B + 2×B – 3×C = 3×C + B + 2×B – 3×C which becomes 2×C = 3×B Divide both sides by 3: 2×C ÷ 3 = 3×B ÷ 3 which makes ⅔×C = B


  

Hint #8


Substitute ⅔×C for B in eq.3d: F = 3×C – ⅔×C which makes F = 2⅓×C


  

Solution

Substitute 3×C for A, ⅔×C for B, 0 for D, 2×C for E, and 2⅓×C for F in eq.1: 3×C + ⅔×C + C + 0 + 2×C + 2⅓×C = 27 which simplifies to 9×C = 27 Divide both sides of the above equation by 9: 9×C ÷ 9 = 27 ÷ 9 which means C = 3 making A = 3×C = 3 × 3 = 9 B = ⅔×C = ⅔ × 3 = 2 E = 2×C = 2 × 3 = 6 F = 2⅓×C = 2⅓ × 3 = 7 and ABCDEF = 923067