Puzzle for July 29, 2023  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 30 eq.2) B + E = C + D eq.3) C + E = D + F - C eq.4) A + B = C + D - B eq.5) C - D - F = D - A - E eq.6)* AB = C + DE

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  AB and DE are 2-digit numbers (not A×B or D×E).

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Hint #1


In eq.4, replace C + D with B + E (from eq.2): A + B = B + E - B which becomes eq.4a) A + B = E


  

Hint #2


Add C to both sides of eq.3: C + E + C = D + F - C + C which becomes eq.3a) 2×C + E = D + F   eq.5 may be written as: eq.5a) C - (D + F) = D - A - E


  

Hint #3


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


  

Hint #4


In eq.4, substitute D + C for A (from eq.5b): D + C + B = C + D - B Subtract D, C, and B from both sides of the above equation: D + C + B - D - C - B = C + D - B - D - C - B which makes 0 = -2×B which means 0 = B


  

Hint #5


Substitute 0 for B in eq.4a: A + 0 = E which makes A = E   eq.6 may be written as: eq.6a) 10×A + B = C + 10×D + E


  

Hint #6


Substitute 0 for B, and A for E in eq.6a: 10×A + 0 = C + 10×D + A which becomes 10×A = C + 10×D + A Subtract A from each side of the above equation: 10×A - A = C + 10×D + A - A which becomes 9×A = C + 10×D which may be written as eq.6b) 9×A = D + C + 9×D


  

Hint #7


Substitute A for D + C (from eq.5b) in eq.6b: 9×A = A + 9×D Subtract A from each side of the equation above: 9×A - A = A + 9×D - A which becomes 8×A = 9×D Divide both sides by 8: 8×A ÷ 8 = 9×D ÷ 8 which makes A = 1⅛×D and also makes E = A = 1⅛×D


  

Hint #8


Substitute 1⅛×D for A in eq.5b: 1⅛×D = D + C Subtract D from both sides of the above equation: 1⅛×D - D = D + C - D which makes ⅛×D = C


  

Hint #9


Substitute ⅛×D for C, and 1⅛×D for E in eq.3: ⅛×D + 1⅛×D = D + F - ⅛×D which becomes 1¼×D = F + ⅞×D Subtract ⅞×D from both sides of the equation above: 1¼×D - ⅞×D = F + ⅞×D - ⅞×D which makes ⅜×D = F


  

Solution

Substitute 1⅛×D for A and E, 0 for B, ⅛×D for C, and ⅜×D for F in eq.1: 1⅛×D + 0 + ⅛×D + D + 1⅛×D + ⅜×D = 30 which simplifies to 3¾×D = 30 Divide both sides of the above equation by 3¾: 3¾×D ÷ 3¾ = 30 ÷ 3¾ which means D = 8 making A = E = 1⅛×D = 1⅛ × 8 = 9 C = ⅛×D = ⅛ × 8 = 1 F = ⅜×D = ⅜ × 8 = 3 and ABCDEF = 901893