Puzzle for March 17, 2020  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) C + D + E = A eq.3) E + F = B eq.4) B + E = C + F eq.5) F – E = A + E eq.6) D + E = A + C – D – E

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 B with E + F (from eq.3): E + F + E = C + F which becomes 2×E + F = C + F Subtract F from both sides of the equation above: 2×E + F – F = C + F – F which makes 2×E = C


  

Hint #2


In eq.2, replace C with 2×E: 2×E + D + E = A which becomes eq.2a) 3×E + D = A


  

Hint #3


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


  

Hint #4


Substitute 3×E for D in eq.2a: 3×E + 3×E = A which makes 6×E = A


  

Hint #5


Substitute 6×E for A in eq.5: F – E = 6×E + E which becomes F – E = 7×E Add E to each side of the above equation: F – E + E = 7×E + E which makes F = 8×E


  

Hint #6


Substitute 8×E for F in eq.3: E + 8×E = B which means 9×E = B


  

Solution

Substitute 6×E for A, 9×E for B, 2×E for C, 3×E for D, and 8×E for F in eq.1: 6×E + 9×E + 2×E + 3×E + E + 8×E = 29 which simplifies to 29×E = 29 Divide both sides of the equation above by 29: 29×E ÷ 29 = 29 ÷ 29 which means E = 1 making A = 6×E = 6 × 1 = 6 B = 9×E = 9 × 1 = 9 C = 2×E = 2 × 1 = 2 D = 3×E = 3 × 1 = 3 F = 8×E = 8 × 1 = 8 and ABCDEF = 692318