Puzzle for August 18, 2020  ( )

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

eq.1) A + B + C + D + E + F = 21 eq.2) D + F = A eq.3) B = A + F eq.4) F – D = C – B eq.5) A + C – E = E – B – F eq.6) B – C = F – B

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

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Hint #1


Subtract F from both sides of eq.3: B – F = A + F – F which becomes B – F = A In eq.2, replace A with B – F: D + F = B – F Add F to each side of the above equation: D + F + F = B – F + F which becomes eq.2a) D + 2×F = B


  

Hint #2


In eq.4, substitute (D + 2×F) for B (from eq.2a): F – D = C – (D + 2×F) which becomes F – D = C – D – 2×F Add D and 2×F to both sides of the above equation: F – D + D + 2×F = C – D – 2×F + D + 2×F which makes 3×F = C


  

Hint #3


Substitute 3×F for C in eq.6: B – 3×F = F – B Add 3×F and B to both sides of the equation above: B – 3×F + 3×F + B = F – B + 3×F + B which becomes 2×B = 4×F Divide both sides by 2: 2×B ÷ 2 = 4×F ÷ 2 which makes B = 2×F


  

Hint #4


Substitute B for 2×F in eq.2a: D + B = B Subtract B from each side of the equation above: D + B – B = B – B which makes D = 0


  

Hint #5


Substitute 0 for D in eq.2: 0 + F = A which makes F = A


  

Hint #6


Substitute F for A, 3×F for C, and 2×F for B in eq.5: F + 3×F – E = E – 2×F – F which becomes 4×F – E = E – 3×F Add E and 3×F to both sides of the above equation: 4×F – E + E + 3×F = E – 3×F + E + 3×F which becomes 7×F = 2×E Divide both sides by 2: 7×F ÷ 2 = 2×E ÷ 2 which makes 3½×F = E


  

Solution

Substitute F for A, 2×F for B, 3×F for C, 0 for D, and 3½×F for E in eq.1: F + 2×F + 3×F + 0 + 3½×F + F = 21 which simplifies to 10½×F = 21 Divide both sides of the equation above by 10½: 10½×F ÷ 10½ = 21 ÷ 10½ which means F = 2 making A = F = 2 B = 2×F = 2 × 2 = 4 C = 3×F = 3 × 2 = 6 E = 3½×F = 3½ × 2 = 7 and ABCDEF = 246072