Puzzle for June 19, 2020  ( )

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

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 the right-side F with A + E (from eq.3): B + C – F = A – E + A + E which becomes B + C – F = 2×A Add F to both sides of the above equation: B + C – F + F = 2×A + F which becomes eq.4a) B + C = 2×A + F


  

Hint #2


In eq.6, replace B + C with 2×A + F (from eq.4a): 2×A + F – D = A + D + F Subtract A and F from both sides of the above equation: 2×A + F – D – A – F = A + D + F – A – F which becomes A – D = D Add D to both sides: A – D + D = D + D which makes A = 2×D


  

Hint #3


In eq.2, substitute 2×D for A: 2×D + B = C + D + E Subtract D from each side of the above equation: 2×D + B – D = C + D + E – D which becomes D + B = C + E which may be written as eq.2a) B + D = C + E


  

Hint #4


Substitute C + E for B + D (from eq.2a) in eq.5: C = C + E Subtract C from both sides of the equation above: C – C = C + E – C which means 0 = E


  

Hint #5


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


  

Hint #6


Substitute B + D for C, and 2×D for A and F in eq.6: B + B + D – D = 2×D + D + 2×D which becomes 2×B = 5×D Divide both sides of the equation above by 2: 2×B ÷ 2 = 5×D ÷ 2 which makes B = 2½×D


  

Hint #7


Substitute 2½×D for B in eq.5: C = 2½×D + D which makes C = 3½×D


  

Solution

Substitute 2×D for A and F, 2½×D for B, 3½×D for C, and 0 for E in eq.1: 2×D + 2½×D + 3½×D + D + 0 + 2×D = 22 which simplifies to 11×D = 22 Divide both sides of the equation above by 11: 11×D ÷ 11 = 22 ÷ 11 which means D = 2 making A = F = 2×D = 2 × 2 = 4 B = 2½×D = 2½ × 2 = 5 C = 3½×D = 3½ × 2 = 7 and ABCDEF = 457204