Puzzle for April 1, 2022  ( )

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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 = A + B eq.3) B + E = A + C + F eq.4) E + F = B + D – F eq.5) D + F – B = B + C – F eq.6) A – D + E = 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.6, replace C + D with A + B (from eq.2): A – D + E = A + B – F Subtract A from each side of the equation above: A – D + E – A = A + B – F – A which becomes –D + E = B – F which is the same as eq.6a) E – D = B – F


  

Hint #2


Subtract F and D from each side of eq.4: E + F – F – D = B + D – F – F – D which becomes eq.4a) E – D = B – 2×F


  

Hint #3


In eq.6a, replace E – D with B – 2×F (from eq.4a): B – 2×F = B – F In the above equation, subtract B from both sides, and add 2×F to both sides: B – 2×F – B + 2×F = B – F – B + 2×F which becomes 0 = F


  

Hint #4


In eq.4, substitute 0 for F: E + 0 = B + D – 0 which becomes eq.4b) E = B + D


  

Hint #5


In eq.5, substitute 0 for F: D + 0 – B = B + C – 0 which becomes D – B = B + C Add B to both sides of the above equation: D – B + B = B + C + B which becomes eq.5a) D = 2×B + C


  

Hint #6


Substitute 2×B + C for D (from eq.5a) in eq.4b: E = B + 2×B + C which becomes eq.4c) E = 3×B + C


  

Hint #7


Substitute 3×B + C for E (from eq.4c), and 0 for F in eq.3: B + 3×B + C = A + C + 0 which becomes 4×B + C = A + C Subtract C from each side of the above equation: 4×B + C – C = A + C – C which makes 4×B = A


  

Hint #8


Substitute 4×B for A, (2×B + C) for D (from eq.5a), 3×B + C for E (from eq.4c), and 0 for F in eq.6: 4×B – (2×B + C) + 3×B + C = C + (2×B + C) – 0 which becomes 7×B – 2×B – C + C = 2×C + 2×B which becomes 5×B = 2×C + 2×B Subtract 2×B from each side of the equation above: 5×B – 2×B = 2×C + 2×B – 2×B which makes 3×B = 2×C Divide both sides by 2: 3×B ÷ 2 = 2×C ÷ 2 which makes 1½×B = C


  

Hint #9


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


  

Hint #10


Substitute 1½×B for C in eq.4c: E = 3×B + 1½×B which makes E = 4½×B


  

Solution

Substitute 4×B for A, 1½×B for C, 3½×B for D, 4½×B for E, and 0 for F in eq.1: 4×B + B + 1½×B + 3½×B + 4½×B + 0 = 29 which simplifies to 14½×B = 29 Divide both sides of the above equation by 14½: 14½×B ÷ 14½ = 29 ÷ 14½ which means B = 2 making A = 4×B = 4 × 2 = 8 C = 1½×B = 1½ × 2 = 3 D = 3½×B = 3½ × 2 = 7 E = 4½×B = 4½ × 2 = 9 and ABCDEF = 823790