Puzzle for July 7, 2020  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) B + C = A eq.3) F = C + E eq.4) E + F = C + D eq.5) A – D = B – E eq.6) C + F – B = A + B – D

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

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Hint #1


In eq.4, replace F with C + E (from eq.3): E + C + E = C + D which becomes C + 2×E = C + D Subtract C from each side of the above equation: C + 2×E – C = C + D – C which makes 2×E = D


  

Hint #2


In eq.5, replace A with B + C (from eq.2), and D with 2×E: B + C – 2×E = B – E In the above equation, subtract B from both sides, and add 2×E to both sides: B + C – 2×E – B + 2×E = B – E – B + 2×E which simplifies to C = E


  

Hint #3


In eq.3, substitute E for C: F = E + E which makes F = 2×E


  

Hint #4


Substitute B + C for A (from eq.2) in eq.6: C + F – B = B + C + B – D In the equation above, subtract C from both sides, and add B to both sides: C + F – B – C + B = B + C + B – D – C + B which simplifies to eq.6a) F = 3×B – D


  

Hint #5


Substitute 2×E for D and F in eq.6a: 2×E = 3×B – 2×E Add 2×E to both sides of the equation above: 2×E + 2×E = 3×B – 2×E + 2×E which makes 4×E = 3×B Divide both sides by 3: 4×E ÷ 3 = 3×B ÷ 3 which makes 1⅓×E = B


  

Hint #6


Substitute 1⅓×E for B, and E for C in eq.2: 1⅓×E + E = A which makes 2⅓×E = A


  

Solution

Substitute 2⅓×E for A, 1⅓×E for B, E for C, and 2×E for D and F in eq.1: 2⅓×E + 1⅓×E + E + 2×E + E + 2×E = 29 which simplifies to 9⅔×E = 29 Divide both sides of the equation above by 9⅔: 9⅔×E ÷ 9⅔ = 29 ÷ 9⅔ which means E = 3 making A = 2⅓×E = 2⅓ × 3 = 7 B = 1⅓×E = 1⅓ × 3 = 4 C = E = 3 D = F = 2×E = 2 × 3 = 6 and ABCDEF = 743636