Puzzle for September 18, 2019  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 26 eq.2) B = D – A – C eq.3) C + E = B + D eq.4) D + F = A + B + C eq.5) E – C – F = B + C eq.6) A = (C + D) ÷ C

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

Scratchpad

 

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Hint #1


Add A and C to both sides of eq.2: B + A + C = D – A – C + A + C which becomes B + A + C = D which may be written as A + B + C = D In eq.4, replace A + B + C with D: D + F = D Subtract D from both sides of the above equation: D + F – D = D – D which means F = 0


  

Hint #2


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


  

Hint #3


In eq.3, substitute B + 2×C for E (from eq.5a): C + B + 2×C = B + D which becomes B + 3×C = B + D Subtract B from both sides of the above equation: B + 3×C – B = B + D – B which makes 3×C = D


  

Hint #4


Substitute 3×C for D in eq.6: A = (C + 3×C) ÷ C which becomes A = (4×C) ÷ C which makes A = 4


  

Hint #5


Substitute 3×C for D, and 4 for A in eq.2: B = 3×C – 4 – C which makes eq.2a) B = 2×C – 4


  

Hint #6


Substitute 2×C – 4 for B (from eq.2a) in eq.5a: E = 2×C – 4 + 2×C which makes eq.5b) E = 4×C – 4


  

Solution

Substitute 4 for A, 2×C – 4 for B (from eq.2a), 3×C for D, 4×C – 4 for E (from eq.5b), and 0 for F in eq.1: 4 + 2×C – 4 + C + 3×C + 4×C – 4 + 0 = 26 which simplifies to 10×C – 4 = 26 Add 4 to both sides of the equation above: 10×C – 4 + 4 = 26 + 4 which becomes 10×C = 30 Divide both sides by 10: 10×C ÷ 10 = 30 ÷ 10 which means C = 3 making B = 2×C – 4 = 2×3 – 4 = 6 – 4 = 2 (from eq.2a) D = 3×C = 3×3 = 9 E = 4×C – 4 = 4×3 – 4 = 12 – 4 = 8 (from eq.5b) and ABCDEF = 423980