Puzzle for June 2, 2021  ( )

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

eq.1) A + B + C + D + E + F = 23 eq.2) B = C + D eq.3) F = B + D eq.4) C + E + F = A eq.5) D + E + F = B + C eq.6) A + C – F = D + E + F

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

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Hint #1


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


  

Hint #2


Multiply both sides of eq.2 by 2: 2×B = 2×(C + D) which becomes 2×B = 2×C + 2×D In the equation above, replace 2×C with D + F (from eq.6a): 2×B = D + F + 2×D which becomes eq.2a) 2×B = 3×D + F


  

Hint #3


In eq.2a, substitute B + D for F (from eq.3): 2×B = 3×D + B + D which becomes 2×B = 4×D + B Subtract B from each side of the above equation: 2×B – B = 4×D + B – B which makes B = 4×D


  

Hint #4


Substitute (4×D) for B in eq.2a: 2×(4×D) = 3×D + F which becomes 8×D = 3×D + F Subtract 3×D from both sides of the equation above: 8×D – 3×D = 3×D + F – 3×D which makes 5×D = F


  

Hint #5


Substitute 4×D for B in eq.2: 4×D = C + D Subtract D from each side of the equation above: 4×D – D = C + D – D which becomes 3×D = C


  

Hint #6


Substitute 5×D for F, 4×D for B, and 3×D for C in eq.5: D + E + 5×D = 4×D + 3×D which becomes 6×D + E = 7×D Subtract 6×D from both sides of the equation above: 6×D + E – 6×D = 7×D – 6×D which makes E = D


  

Hint #7


Substitute 3×D for C, D for E, and 5×D for F in eq.4: 3×D + D + 5×D = A which becomes 9×D = A


  

Solution

Substitute 9×D for A, 4×D for B, 3×D for C, D for E, and 5×D for F in eq.1: 9×D + 4×D + 3×D + D + D + 5×D = 23 which simplifies to 23×D = 23 Divide both sides of the above equation by 23: 23×D ÷ 23 = 23 ÷ 23 which means D = 1 making A = 9×D = 9 × 1 = 9 B = 4×D = 4 × 1 = 4 C = 3×D = 3 × 1 = 3 E = D = 1 F = 5×D = 5 × 1 = 5 and ABCDEF = 943115