Puzzle for December 15, 2022  ( )

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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 + E = B + D eq.3) E - F = A - C eq.4) D - C = C - F eq.5)* BC = D + E + F eq.6) E = average (A, B, D, F)

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  BC is a 2-digit number (not B×C).

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Hint #1


Add F and C to both sides of eq.3: E - F + F + C = A - C + F + C which becomes eq.3a) E + C = A + F


  

Hint #2


eq.6 may be written as: E = (A + B + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (A + B + D + F) ÷ 4 which becomes eq.6a) 4×E = A + B + D + F which may be written as eq.6b) 4×E = A + F + B + D


  

Hint #3


In eq.6b, replace A + F with E + C (from eq.3a), and B + D with C + E (from eq.2): 4×E = E + C + C + E which becomes 4×E = 2×E + 2×C Subtract 2×E from each side of the equation above: 4×E - 2×E = 2×E + 2×C - 2×E which makes 2×E = 2×C Divide both sides by 2: 2×E ÷ 2 = 2×C ÷ 2 which makes E = C


  

Hint #4


eq.5 may be re-written as: 10×B + C = D + E + F In the equation above, substitute C for E: 10×B + C = D + C + F Subtract C from both sides: 10×B + C - C = D + C + F - C which becomes eq.5a) 10×B = D + F


  

Hint #5


Add C and F to both sides of eq.4: D - C + C + F = C - F + C + F which becomes eq.4a) D + F = 2×C


  

Hint #6


Substitute 2×C for D + F (from eq.4a) into eq.5a: 10×B = 2×C Divide both sides of the above equation by 2: 10×B ÷ 2 = 2×C ÷ 2 which makes 5×B = C and also makes 5×B = C = E


  

Hint #7


Substitute 5×B for C and E in eq.2: 5×B + 5×B = B + D which becomes 10×B = B + D Subtract B from both sides of the above equation: 10×B - B = B + D - B which makes 9×B = D


  

Hint #8


Substitute 9×B for D in eq.5a: 10×B = 9×B + F Subtract 9×B from each side of the equation above: 10×B - 9×B = 9×B + F - 9×B which makes B = F


  

Hint #9


Substitute (5×B) for E, and 10×B for D + F (from eq.5a) in eq.6a: 4×(5×B) = A + B + 10×B which becomes 20×B = A + 11×B Subtract 11×B from each side of the above equation: 20×B - 11×B = A + 11×B - 11×B which makes 9×B = A


  

Solution

Substitute 9×B for A and D, 5×B for C and E, and B for F in eq.1: 9×B + B + 5×B + 9×B + 5×B + B = 30 which simplifies to 30×B = 30 Divide both sides of the above equation by 30: 30×B ÷ 30 = 30 ÷ 30 which means B = 1 making A = D = 9×B = 9 × 1 = 9 C = E = 5×B = 5 × 1 = 5 F = B = 1 and ABCDEF = 915951