Puzzle for February 16, 2019  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 29 eq.2) B + D = A - D eq.3) A + B = F eq.4) F - D = D + E - A - B eq.5)* EF - D = AB + D eq.6)* CD - EF = A + B + C + D

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

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Hint #1


Add (D - B) to both sides of eq.2: B + D + (D - B) = A - D + (D - B) which becomes eq.2a) 2×D = A - B


  

Hint #2


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


  

Hint #3


In eq.4a, replace F with A + B (from eq.3): A + B = E - 2×B Add 2×B to each side of the above equation: A + B + 2×B = E - 2×B + 2×B which becomes eq.4b) A + 3×B = E


  

Hint #4


eq.5 may be stated as: 10×E + F - D = 10×A + B + D Add D to both sides of the above equation: 10×E + F - D + D = 10×A + B + D + D which becomes eq.5a) 10×E + F = 10×A + B + 2×D


  

Hint #5


Substitute (A + 3×B) for E (from eq.4a), A + B for F (from eq.3), and A - B for 2×D (from eq.2a) in eq.5a: 10×(3×B + A) + A + B = 10×A + B + A - B which becomes 30×B + 10×A + A + B = 11×A which becomes 31×B + 11×A = 11×A Subtract 11×A from both sides: 31×B + 11×A - 11×A = 11×A - 11×A which makes 31×B = 0 which means B = 0


  

Hint #6


Substitute 0 for B in eq.2a: 2×D = A - 0 which means 2×D = A


  

Hint #7


Substitute 2×D for A, and 0 for B in eq.4b: 2×D + 3×0 = E which means 2×D = E


  

Hint #8


Substitute 2×D for A, and 0 for B in eq.3: 2×D + 0 = F which makes 2×D = F


  

Hint #9


eq.6 may be written as: 10×C + D - (10×E + F) = A + B + C + D which is the same as 10×C + D - 10×E - F = A + B + C + D Subtract both C and D from each side of the above equation: 10×C + D - 10×E - F - C - D = A + B + C + D - C - D which becomes eq.6a) 9×C - 10×E - F = A + B


  

Hint #10


Substitute 0 for B, and 2×D for A and E and F in eq.6a: 9×C - 10×2×D - 2×D = 2×D + 0 which is the same as 9×C - 20×D - 2×D = 2×D which becomes 9×C - 22×D = 2×D Addd 22×D to each side: 9×C - 22×D + 22×D = 2×D + 22×D which makes 9×C = 24×D Divide both sides by 9: 9×C ÷ 9 = 24×D ÷ 9 which makes C = 2⅔×D


  

Solution

Substitute 2×D for A and E and F, 0 for B, and 2⅔×D for C in eq.1: 2×D + 0 + 2⅔×D + D + 2×D + 2×D = 29 which simplifies to 9⅔×D = 29 Divide each side by 9⅔: 9⅔×D ÷ 9⅔ = 29 ÷ 9⅔ which makes D = 3 making A = E = F = 2×D = 2 × 3 = 6 C = 2⅔×D = 2⅔ × 3 = 8 and ABCDEF = 608366