Puzzle for February 25, 2020  ( )

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

eq.1) A + B + C + D + E + F = 23 eq.2) D + E – B = A + B + C eq.3) B + C = A – C eq.4) C + E = A + F eq.5) A + B = E – B – C eq.6) E – C = A + 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.2, replace A + B with E – B – C (from eq.5): D + E – B = E – B – C + C which becomes D + E – B = E – B In the above equation, subtract E from both sides, and add B to both sides: D + E – B – E + B = E – B – E + B which simplifies to D = 0


  

Hint #2


eq.5 may be written as: A + B = E – C – B In the equation above, replace E – C with A + C + D (from eq.6): A + B = A + C + D – B Add B to both sides, subtract A from each side, and replace D with 0: A + B + B – A = A + C + 0 – B + B – A which simplifies to 2×B = C


  

Hint #3


In eq.3, substitute 2×B for C: B + 2×B = A – 2×B which becomes 3×B = A – 2×B Add 2×B to both sides of the above equation: 3×B + 2×B = A – 2×B + 2×B which makes 5×B = A


  

Hint #4


Substitute 2×B for C, 5×B for A, and 0 for D in eq.6: E – 2×B = 5×B + 2×B + 0 which becomes E – 2×B = 7×B Add 2×B to each side of the above equation: E – 2×B + 2×B = 7×B + 2×B which makes E = 9×B


  

Hint #5


Substitute 2×B for C, 9×B for E, and 5×B for A in eq.4: 2×B + 9×B = 5×B + F which becomes 11×B = 5×B + F Subtract 5×B from each side of the equation above: 11×B – 5×B = 5×B + F – 5×B which makes 6×B = F


  

Solution

Substitute 5×B for A, 2×B for C, 0 for D, 9×B for E, and 6×B for F in eq.1: 5×B + B + 2×B + 0 + 9×B + 6×B = 23 which simplifies to 23×B = 23 Divide both sides of the above equation by 23: 23×B ÷ 23 = 23 ÷ 23 which means B = 1 making A = 5×B = 5 × 1 = 5 C = 2×B = 2 × 1 = 2 E = 9×B = 9 × 1 = 9 F = 6×B = 6 × 1 = 6 and ABCDEF = 512096