Puzzle for October 19, 2022  ( )

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

eq.1) A + B + C + D + E + F = 22 eq.2) F = D + E eq.3) E - C = A - E eq.4) A - C = B + C eq.5) E + F = B - C + D eq.6) C + D + F = A - D + E

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

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Hint #1


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


  

Hint #2


In eq.4, replace B with 2×E + C (from eq.5a): A - C = 2×E + C + C which becomes A - C = 2×E + 2×C Add C to both sides of the above equation: A - C + C = 2×E + 2×C + C which becomes eq.4a) A = 2×E + 3×C


  

Hint #3


In eq.3, substitute 2×E + 3×C for A (from eq.4a): E - C = 2×E + 3×C - E which becomes E - C = E + 3×C In the equation above, subtract E from both sides, and add C to both sides: E - C - E + C = E + 3×C - E + C which makes 0 = 4×C which means 0 = C


  

Hint #4


Substitute 0 for C in eq.4a: A = 2×E + 3×0 which becomes A = 2×E + 0 which makes A = 2×E


  

Hint #5


Substitute 0 for C in eq.5a: 2×E + 0 = B which makes 2×E = B


  

Hint #6


Substitute 0 for C, D + E for F (from eq.2), and 2×E for A in eq.6: 0 + D + D + E = 2×E - D + E which makes 2×D + E = 3×E - D In the above equation, subtract E from both sides, and add D to both sides: 2×D + E - E + D = 3×E - D - E + D which makes 3×D = 2×E Divide both sides by 3: 3×D ÷ 3 = 2×E ÷ 3 which makes D = ⅔×E


  

Hint #7


Substitute ⅔×E for D in eq.2: F = ⅔×E + E which makes F = 1⅔×E


  

Solution

Substitute 2×E for A and B, 0 for C, ⅔×E for D, and 1⅔×E for F in eq.1: 2×E + 2×E + 0 + ⅔×E + E + 1⅔×E = 22 which simplifies to 7⅓×E = 22 Divide both sides of the above equation by 7⅓: 7⅓×E ÷ 7⅓ = 22 ÷ 7⅓ which means E = 3 making A = B = 2×E = 2 × 3 = 6 D = ⅔×E = ⅔ × 3 = 2 F = 1⅔×E = 1⅔ × 3 = 5 and ABCDEF = 660235