Puzzle for June 30, 2021  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) C = A + F eq.3) E + F = B eq.4) B – D = D + E eq.5) B – C + D = E – D eq.6) D + E = C – E + F

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

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Hint #1


Add D to both sides of eq.4: B – D + D = D + E + D which beomes eq.4a) B = 2×D + E   In eq.3, replace B with 2×D + E (from eq.4a): E + F = 2×D + E Subtract E from both sides of the above equation: E + F – E = 2×D + E – E which makes F = 2×D


  

Hint #2


Add the left and right sides of eq.5 to the left and right sides of eq.4, respectively: B – D + (B – C + D) = D + E + (E – D) which becomes 2×B – C = 2×E In the equation above, add C to both sides, and subtract 2×E from both sides: 2×B – C + C – 2×E = 2×E + C – 2×E which becomes 2×B – 2×E = C which may be written as eq.4b) 2×(B – E) = C


  

Hint #3


Subtract E from both sides of eq.4a: B – E = 2×D + E – E which makes B – E = 2×D In eq.4b, replace B – E with 2×D: 2×(2×D) = C which makes 4×D = C


  

Hint #4


In eq.2, substitute 4×D for C, and 2×D for F: 4×D = A + 2×D Subtract 2×D from each side of the equation above: 4×D – 2×D = A + 2×D – 2×D which becomes 2×D = A


  

Hint #5


In eq.6, substitute 4×D for C, and 2×D for F: D + E = 4×D – E + 2×D which becomes D + E = 6×D – E In the above equation, add E to both sides, and subtract D from both sides: D + E + E – D = 6×D – E + E – D which becomes 2×E = 5×D Divide both sides by 2: 2×E ÷ 2 = 5×D ÷ 2 which makes E = 2½×D


  

Hint #6


Substitute 2½×D for E, and 2×D for F in eq.3: 2½×D + 2×D = B which makes 4½×D = B


  

Solution

Substitute 2×D for A and F, 4½×D for B, 4×D for C, and 2½×D for E in eq.1: 2×D + 4½×D + 4×D + D + 2½×D + 2×D = 32 which simplifies to 16×D = 32 Divide both sides of the above equation by 16: 16×D ÷ 16 = 32 ÷ 16 which means D = 2 making A = F = 2×D = 2 × 2 = 4 B = 4½×D = 4½ × 2 = 9 C = 4×D = 4 × 2 = 8 E = 2½×D = 2½ × 2 = 5 and ABCDEF = 498254