Puzzle for July 24, 2019  ( )

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

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: A - B + B + C = F - C + B + C which becomes A + C = F + B In eq.6, replace A + C with F + B: D + F = F + B Subtract F from both sides of the above equation: D + F - F = F + B - F which makes D = B


  

Hint #2


In eq.4, replace D with B: F = A + B + B which becomes eq.4a) F = A + 2×B


  

Hint #3


In eq.6, substitute B for D, and A + 2×B for F (from eq.4a): B + A + 2×B = A + C Subtract A from both sides of the equation above: B + A + 2×B - A = A + C - A which makes 3×B = C


  

Hint #4


Substitute 3×B for C, and B for D in eq.2: 3×B - B = B + E Subtract B from both sides: 3×B - B - B = B + E - B which makes B = E


  

Hint #5


Substitute 3×B for C, B for D and E, and A + 2×B for F (from eq.4a) in eq.5: 3×B + B + B = A + A + 2×B which becomes 5×B = 2×A + 2×B Subtract 2×B from each side: 5×B - 2×B = 2×A + 2×B - 2×B which means 3×B = 2×A Divide both sides by 2: 3×B ÷ 2 = 2×A ÷ 2 which makes 1½×B = A


  

Hint #6


Substitute 1½×B for A in eq.4a: F = 1½×B + 2×B which makes F = 3½×B


  

Solution

Substitute 1½×B for A, 3×B for C, B for D and E, and 3½×B for F in eq.1: 1½×B + B + 3×B + B + B + 3½×B = 22 which simplifies to 11×B = 22 Divide both sides of the equation above by 11: 11×B ÷ 11 = 22 ÷ 11 which means B = 2 making A = 1½×B = 1½ × 2 = 3 C = 3×B = 3 × 2 = 6 D = E = B = 2 F = 3½×B = 3½ × 2 = 7 and ABCDEF = 326227