Puzzle for January 15, 2020  ( )

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

eq.1) A + B + C + D + E + F = 17 eq.2) D + E = A + B + C eq.3) E = A + F eq.4) B – C = E – D eq.5) F = C + D eq.6) B = A + D

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


  

Hint #2


In eq.2a, replace F with C + D (from eq.5): D + C + D = B + C which becomes C + 2×D = B + C Subtract C from both sides of the above equation: C + 2×D – C = B + C – C which makes eq.2b) 2×D = B


  

Hint #3


In eq.2b, substitute A + D for B (from eq.6): 2×D = A + D Subtract D from both sides of the above equation: 2×D – D = A + D – D which makes D = A


  

Hint #4


Substitute D for A, and C + D for F (from eq.5) in eq.3: E = D + C + D which becomes eq.3a) E = C + 2×D


  

Hint #5


Substitute 2×D for B, and C + 2×D for E (from eq.3a) in eq.4: 2×D – C = C + 2×D – D which becomes 2×D – C = C + D In the equation above, add C to each side, and subtract D from each side: 2×D – C + C – D = C + D + C – D which makes D = 2×C and also makes A = D = 2×C


  

Hint #6


Substitute (2×C) for D in eq.2b: 2×(2×C) = B which makes 4×C = B


  

Hint #7


In eq.3a, substitute (2×C) for D: E = C + 2×(2×C) which becomes E = C + 4×C which makes E = 5×C


  

Hint #8


Substitute 2×C for D in eq.5: F = C + 2×C which makes F = 3×C


  

Solution

Substitute 2×C for A and D, 4×C for B, 5×C for E, and 3×C for F in eq.1: 2×C + 4×C + C + 2×C + 5×C + 3×C = 17 which simplifies to 17×C = 17 Divide both sides of the equation above by 17: 17×C ÷ 17 = 17 ÷ 17 which means C = 1 making A = D = 2×C = 2 × 1 = 2 B = 4×C = 4 × 1 = 4 E = 5×C = 5 × 1 = 5 F = 3×C = 3 × 1 = 3 and ABCDEF = 241253