Puzzle for September 10, 2020  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) A + B + C = D + E + F eq.3) D + F = A – B eq.4) B + E = C – D eq.5)* AB + C – (A – C) = CD – C

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  AB and CD are 2-digit numbers (not A×B or C×D).

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Hint #1


Subtract D and E from both sides of eq.2: A + B + C – D – E = D + E + F – D – E which becomes A + B + C – D – E = F In the above equation, replace C – D with B + E (from eq.4): A + B + B + E – E = F which becomes eq.2a) A + 2×B = F


  

Hint #2


In eq.3, replace F with A + 2×B (from eq.2a): D + A + 2×B = A – B In the above equation, subtract A from both sides, and add B to both sides: D + A + 2×B – A + B = A – B – A + B which becomes D + 3×B = 0 which makes D = B = 0


  

Hint #3


In eq.2a, substitute 0 for B: A + 2×0 = F which makes A = F


  

Hint #4


Substitute 0 for B and D in eq.4: 0 + E = C – 0 which makes E = C


  

Hint #5


eq.5 may be written as: 10×A + B + C – A + C = 10×C + D – C which becomes 9×A + B + 2×C = 9×C + D Subtract 2×C from both sides of the above equation: 9×A + B + 2×C – 2×C = 9×C + D – 2×C which becomes eq.5a) 9×A + B = 7×C + D


  

Hint #6


Substitute 0 for B and D in eq.5a: 9×A + 0 = 7×C + 0 which makes 9×A = 7×C and also makes eq.5b) 9×A = 7×C = 7×E


  

Hint #7


Multiply both sides of eq.1 by 7: 7×(A + B + C + D + E + F) = 7×32 which is the same as 7×A + 7×B + 7×C + 7×D + 7×E + 7×F = 224 Substitute 0 for B and D, 9×A for 7×C and 7×E (from eq.5b), and A for F in the equation above: 7×A + 7×0 + 9×A + 7×0 + 9×A + 7×A = 224 which simplifies to 32×A = 224 Divide both sides by 32: 32×A ÷ 32 = 224 ÷ 32 which means A = 7 and also means F = A = 7


  

Solution

Substitute 7 for A in eq.5b: 9×7 = 7×C = 7×E which makes 63 = 7×C = 7×E Divide all sides of the equation above by 7: 63 ÷ 7 = 7×C ÷ 7 = 7×E ÷ 7 which makes 9 = C = E and makes ABCDEF = 709097