Puzzle for September 14, 2020  ( )

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

eq.1) A + B + C + D + E + F = 30 eq.2) C + F = A + B eq.3) D + F = C + E eq.4) B = C + D eq.5) C + D + E = B + F eq.6) A – B – C – D = B + C + D

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

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Hint #1


In eq.5, replace C + D with B (from eq.4): B + E = B + F Subtract B from both sides of the equation above: B + E – B = B + F – B which makes E = F


  

Hint #2


In eq.3, replace F with E: D + E = C + E Subtract E from both sides of the above equation: D + E – E = C + E – E which makes D = C


  

Hint #3


In eq.4, substitute C for D: B = C + C which makes B = 2×C


  

Hint #4


Substitute 2×C for B, and C for D in eq.6: A – 2×C – C – C = 2×C + C + C which becomes A – 4×C = 4×C Add 4×C to both sides of the equation above: A – 4×C + 4×C = 4×C + 4×C which makes A = 8×C


  

Hint #5


Substitute 8×C for A, and 2×C for B in eq.2: C + F = 8×C + 2×C which becomes C + F = 10×C Subtract C from both sides of the equation above: C + F – C = 10×C – C which makes F = 9×C and also makes E = F = 9×C


  

Solution

Substitute 8×C for A, 2×C for B, C for D, and 9×C for E and F in eq.1: 8×C + 2×C + C + C + 9×C + 9×C = 30 which simplifies to 30×C = 30 Divide both sides of the above equation by 30: 30×C ÷ 30 = 30 ÷ 30 which means C = 1 making A = 8×C = 8 × 1 = 8 B = 2×C = 2 × 1 = 2 D = C = 1 E = F = 9×C = 9 × 1 = 9 and ABCDEF = 821199