Puzzle for December 24, 2018  ( )

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

eq.1) A + B + C + D + E + F = 28 eq.2) C + F = B + D eq.3) A + B = E eq.4) B - D = A - E eq.5) C + D = A eq.6) B + C + E = A + D + F

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

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Hint #1


Add D + E to both sides of eq.4: B - D + D + E = A - E + D + E which becomes eq.4a) B + E = A + D Substitute A + B for E (from eq.3) in eq.4a above: B + A + B = A + D Subtract A from each side of the above equation: B + A + B - A = A + D - A which makes 2×B = D


  

Hint #2


In eq.6, replace A + D with B + E (from eq.4a): B + C + E = B + E + F Subtract both B and E from each side of the above equation: B + C + E - B - E = B + E + F - B - E which makes C = F


  

Hint #3


Substitute 2×B for D, and C for F in eq.2: C + C = B + 2×B which simplifies to 2×C = 3×B Divide both sides of the above equation by 2: 2×C ÷ 2 = 3×B ÷ 2 which means C = 1½×B


  

Hint #4


Substitute 1½×B for C, and 2×B for D in eq.5: 1½×B + 2×B = A which makes 3½×B = A


  

Hint #5


In eq.3, replace A with 3½×B: 3½×B + B = E which makes 4½×B = E


  

Solution

Substitute 3½×B for A, 1½×B for C and F, 2×B for D, and 4½×B for E in eq.1: 3½×B + B + 1½×B + 2×B + 4½×B + 1½×B = 28 which simplifies to 14×B = 28 Divide both sides of the equation above by 14: 14×B ÷ 14 = 28 ÷ 14 which means B = 2 making A = 3½×B = 3½ × 2 = 7 C = F = 1½×B = 1½ × 2 = 3 D = 2×B = 2 × 2 = 4 E = 4½×B = 4½ × 2 = 9 and ABCDEF = 723493