Puzzle for March 10, 2020  ( )

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

eq.1) A + B + C + D + E + F = 31 eq.2) E + F = D eq.3) D – C = A – B eq.4) B + D = C + F eq.5) C + E = A + F eq.6) A + B + C = D + E

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

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Hint #1


Add B and C to both sides of eq.3: D – C + B + C = A – B + B + C which becomes D + B = A + C which may be written as eq.3a) B + D = A + C   In eq.4, replace B + D with A + C (from eq.3a): A + C = C + F Subtract C from both sides of the above equation: A + C – C = C + F – C which makes A = F


  

Hint #2


eq.6 may be re–written as: A + C + B = D + E In the equation above, replace A + C with B + D (from eq.3a): B + D + B = D + E which becomes 2×B + D = D + E Subtract D from each side: 2×B + D – D = D + E – D which makes 2×B = E


  

Hint #3


In eq.2, substitute 2×B for E: eq.2a) 2×B + F = D


  

Hint #4


Substitute 2×B + F for D (from eq.2a) in eq.4: B + 2×B + F = C + F which becomes 3×B + F = C + F Subtract F from each side of the above equation: 3×B + F – F = C + F – F which makes 3×B = C


  

Hint #5


Substitute 3×B for C, 2×B for E, and A for F in eq.5: 3×B + 2×B = A + A which becomes 5×B = 2×A Divide both sides of the equation above by 2: 5×B ÷ 2 = 2×A ÷ 2 which makes 2½×B = A and also makes F = A = 2½×B


  

Hint #6


Substitute 2×B for E, and 2½×B for F in eq.2: 2×B + 2½×B = D which makes 4½×B = D


  

Solution

Substitute 2½×B for A and F, 3×B for C, 4½×B for D, and 2×B for E in eq.1: 2½×B + B + 3×B + 4½×B + 2×B + 2½×B = 31 which simplifies to 15½×B = 31 Divide both sides of the above equation by 15½: 15½×B ÷ 15½ = 31 ÷ 15½ which means B = 2 making A = F = 2½×B = 2½ × 2 = 5 C = 3×B = 3 × 2 = 6 D = 4½×B = 4½ × 2 = 9 E = 2×B = 2 × 2 = 4 and ABCDEF = 526945