Puzzle for June 27, 2020  ( )

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

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

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

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Hint #1


In eq.3, replace B + C with E (from eq.4): eq.3a) A + D = E + F


  

Hint #2


In eq.1, replace E + F with A + D (from eq.3a): B + A + D = A + C + D Subtract A and D from both sides of the above equation: B + A + D – A – D = A + C + D – A – D which simplifies to B = C


  

Hint #3


In eq.4, substitute B for C: B + B = E which makes eq.4a) 2×B = E


  

Hint #4


Substitute A + E for D (from eq.2), and B for C in eq.3: A + A + E = B + B + F which becomes eq.3b) 2×A + E = 2×B + F


  

Hint #5


Substitute E for 2×B in eq.3b: 2×A + E = E + F Subtract E from each side of the equation above: 2×A + E – E = E + F – E which makes 2×A = F


  

Hint #6


eq.5 may be written as: 10×E + F – (10×C + D) = 10×A + B + D which is the same as 10×E + F – 10×C – D = 10×A + B + D Add D to each side of the above equation: 10×E + F – 10×C – D + D = D + 10×A + B + D which becomes eq.5a) 10×E + F – 10×C = 10×A + B + 2×D


  

Hint #7


Substitute (2×B) for E, 2×A for F, and B for C in eq.5a: 10×(2×B) + 2×A – 10×B = 2×D + 10×A + B which is the same as 20×B + 2×A – 10×B = 2×D + 10×A + B which becomes 10×B + 2×A = 2×D + 10×A + B Subtract 2×A, B, and 2×D from both sides of the above equation: 10×B + 2×A – 2×A – B – 2×D = 2×D + 10×A + B – B – 2×A – 2×D which simplifies to eq.5b) 9×B – 2×D = 8×A


  

Hint #8


Substitute (A + E) for D (from eq.2) in eq.5b: 9×B – 2×(A + E) = 8×A which is equivalent to 9×B – 2×A – 2×E = 8×A Add 2×A to each side of the equation above: 9×B – 2×A – 2×E + 2×A = 8×A + 2×A which becomes eq.5c) 9×B – 2×E = 10×A


  

Hint #9


Substitute (2×B) for E in eq.5c: 9×B – 2×(2×B) = 10×A which is the same as 9×B – 4×B = 10×A which becomes 5×B = 10×A Divide both sides of the equation above by 5: 5×B ÷ 5 = 10×A ÷ 5 which makes C = 2×A and also makes B = C = 2×A


  

Hint #10


Substitute (2×A) for B in eq.4a: 2×(2×A) = E which makes 4×A = E


  

Hint #11


Substitute 4×A for E in eq.2: D = A + 4×A which makes D = 5×A


  

Solution

Substitute 2×A for both C and B in eq.6: 2×A = A × 2×A Divide both sides of the above equation by 2×A: 2×A ÷ 2×A = A × 2×A ÷ 2×A which makes 1 = A making B = C = F = 2×A = 2 × 1 = 2 D = 5×A = 5 × 1 = 5 E = 4×A = 4 × 1 = 4 and ABCDEF = 122542