Puzzle for May 11, 2022  ( )

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

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

A, B, C, D, E, and F each represent a one-digit non-negative integer.
 
Our thanks go out to Judah S (age 15) for sending us today's puzzle! Thank you for today's puzzle, Judah, and for all of this week's puzzles!

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Hint #1


In eq.3, replace A + B with E – C (from eq.1): D + E = E – C – F In the above equation, subtract E from both sides, and add C and F to both sides: D + E – E + C + F = E – C – F – E + C + F which simplifies to C + D + F = 0 Since C, D, and F must all be non-negative, the above equation makes: C = 0 and D = 0 and F = 0


  

Hint #2


In eq.1, replace C with 0: A + B = E – 0 which becomes eq.1a) A + B = E


  

Hint #3


In eq.2, substitute 0 for D and F, and A + B for E (from eq.1a): B – A – 0 = A + B – B – 0 which becomes B – A = A Add A to both sides of the equation above: B – A + A = A + A which makes B = 2×A


  

Hint #4


Substitute 2×A for B in eq.1a: A + 2×A = E which makes 3×A = E


  

Solution

Substitute (3×A) for E, 2×A for B, and 0 for F in eq.4: (3×A) ÷ A = A + 2×A + 0 which makes 3 = 3×A Divide both sides of the above equation by 3: 3 ÷ 3 = 3×A ÷ 3 which makes 1 = A making B = 2×A = 2 × 1 = 2 E = 3×A = 3 × 1 = 3 and ABCDEF = 120030