Puzzle for May 26, 2020  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) E + F = A eq.3) D + E = B eq.4) B + E = A eq.5) C + E = B + D eq.6) A + B – D = C + D + F

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

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Hint #1


In eq.2, replace A with B + E (from eq.4): E + F = B + E Subtract E from each side of the equation above: E + F – E = B + E – E which makes F = B


  

Hint #2


In eq.5, replace B with D + E (from eq.3): C + E = D + E + D which becomes C + E = 2×D + E Subtract E from each side of the equation above: C + E – E = 2×D + E – E which makes C = 2×D


  

Hint #3


In eq.6, substitute 2×D for C, and B for F: A + B – D = 2×D + D + B which becomes A + B – D = 3×D + B In the above equation, subtract B from both sides, and add D to both sides: A + B – D – B + D = 3×D + B – B + D which makes A = 4×D


  

Hint #4


Substitute D + E for B (from eq.3), and 4×D for A in eq.4: D + E + E = 4×D which becomes D + 2×E = 4×D Subtract D from both sides of the equation above: D + 2×E – D = 4×D – D which makes 2×E = 3×D Divide both sides by 2: 2×E ÷ 2 = 3×D ÷ 2 which makes E = 1½×D


  

Hint #5


Substitute 1½×D for E in eq.3: D + 1½×D = B which makes 2½×D = B and also makes F = B = 2½×D


  

Solution

Substitute 4×D for A, 2½×D for B and F, 2×D for C, and 1½×D for E in eq.1: 4×D + 2½×D + 2×D + D + 1½×D + 2½×D = 27 which simplifies to 13½×D = 27 Divide both sides of the equation above by 13½: 13½×D ÷ 13½ = 27 ÷ 13½ which means D = 2 making A = 4×D = 4 × 2 = 8 B = F = 2½×D = 2½ × 2 = 5 C = 2×D = 2 × 2 = 4 E = 1½×D = 1½ × 2 = 3 and ABCDEF = 854235