Puzzle for September 8, 2020  ( )

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

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

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Hint #1


Add C to both sides of eq.4: B + C + C = A – C + C which becomes eq.4a) B + 2×C = A   In eq.2, substitute (B + 2×C) for A (from eq.4a): C – D = B – (B + 2×C) which becomes C – D = B – B – 2×C which becomes C – D = –2×C Add D and 2×C to both sides of the equation above: C – D + D + 2×C = –2×C + D + 2×C which makes 3×C = D


  

Hint #2


In eq.5, replace D with 3×C: C + 3×C = B – 3×C which becomes 4×C = B – 3×C Add 3×C to both sides of the above equation: 4×C + 3×C = B – 3×C + 3×C which becomes 7×C = B


  

Hint #3


In eq.4a, replace B with 7×C: 7×C + 2×C = A which becomes 9×C = A


  

Hint #4


In eq.2, add A to both sides, and subtract C from both sides: C – D + A – C = B – A + A – C which becomes –D + A = B – C In the above equation, substitute E – D – F for B – C (from eq.3): –D + A = E – D – F Add D to both sides: –D + A + D = E – D – F + D which becomes eq.2a) A = E – F


  

Hint #5


In eq.2a, substitute E + F for A (from eq.6): E + F = E – F In the equation above, subtract E from both sides, and add F to both sides: E + F – E + F = E – F – E + F which makes 2×F = 0 which means F = 0


  

Hint #6


Substitute 9×C for A, and 0 for F in eq.6: 9×C = E + 0 which makes 9×C = E


  

Solution

Substitute 9×C for A and E, 7×C for B, 3×C for D, and 0 for F in eq.1: 9×C + 7×C + C + 3×C + 9×C + 0 = 29 which simplifies to 29×C = 29 Divide both sides of the above equation by 29: 29×C ÷ 29 = 29 ÷ 29 which means C = 1 making A = E = 9×C = 9 × 1 = 9 B = 7×C = 7 × 1 = 7 D = 3×C = 3 × 1 = 3 and ABCDEF = 971390