Puzzle for January 14, 2019  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) A + B = F eq.3) A = B + C eq.4) C + D + E = F eq.5) C + F = B + E eq.6) A + C = B + D

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 A with B + C (from eq.3): B + C + C = B + D which is equivalent to B + 2×C = B + D Subtract B from both sides of the equation above: B + 2×C - B = B + D - B which makes 2×C = D


  

Hint #2


In eq.4, replace D with 2×C: C + 2×C + E = F which becomes 3×C + E = F In eq.5, substitute 3×C + E for F: C + 3×C + E = B + E which becomes 4×C + E = B + E Subtract E from each side of the above equation: 4×C + E - E = B + E - E which makes 4×C = B


  

Hint #3


Substitute 4×C for B, and 2×C for D in eq.6: A + C = 4×C + 2×C which becomes A + C = 6×C Subtract C from each side of the above equation: A + C - C = 6×C - C which makes A = 5×C


  

Hint #4


Substitute 5×C for A, and 4×C for B in eq.2: 5×C + 4×C = F which means 9×C = F


  

Hint #5


Substitute 4×C for B, and 9×C for F in eq.5: C + 9×C = 4×C + E Subtract 4×C from both sides: C + 9×C - 4×C = 4×C + E - 4×C which makes 6×C = E


  

Solution

Substitute 5×C for A, 4×C for B, 2×C for D, 6×C for E, and 9×C for F in eq.1: 5×C + 4×C + C + 2×C + 6×C + 9×C = 27 which simplifies to 27×C = 27 Divide both sides by 27: 27×C ÷ 27 = 27 ÷ 27 which means C = 1 making A = 5×C = 5 × 1 = 5 B = 4×C = 4 × 1 = 4 D = 2×C = 2 × 1 = 2 E = 6×C = 6 × 1 = 6 F = 9×C = 9 × 1 = 9 and ABCDEF = 541269