Puzzle for April 21, 2020  ( )

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

eq.1) A + B + C + D + E + F = 19 eq.2) D + E + F = A + C eq.3) B + D = A + C + E eq.4) B – F = A + E + F eq.5) A – C = D + F eq.6) E – F = C + F

A, B, C, D, E, and F each represent a one-digit non-negative integer.

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Hint #1


eq.2 may be written as: D + F + E = A + C In the above equation, replace D + F with A – C (from eq.5): A – C + E = A + C Subtract A from each side, and add C to both sides: A – C + E – A + C = A + C – A + C which simplifies to eq.2a) E = 2×C


  

Hint #2


In eq.6, replace E with 2×C: 2×C – F = C + F In the equation above, add F to both sides, and subtract C from each side: 2×C – F + F – C = C + F + F – C which makes C = 2×F


  

Hint #3


In eq.2a, substitute (2×F) for C: E = 2×(2×F) which makes E = 4×F


  

Hint #4


Substitute 4×F for E, and 2×F for C in eq.2: D + 4×F + F = A + 2×F which becomes D + 5×F = A + 2×F Subtract 2×F from both sides of the above equation: D + 5×F – 2×F = A + 2×F – 2×F which becomes eq.2b) D + 3×F = A


  

Hint #5


Substitute D + 3×F for A (from eq.2b), 2×F for C, and 4×F for E in eq.3: B + D = D + 3×F + 2×F + 4×F which becomes B + D = D + 9×F Subtract D from both sides of the above equation: B + D – D = D + 9×F – D which makes B = 9×F


  

Hint #6


Substitute 9×F for B, and 4×F for E in eq.4: 9×F – F = A + 4×F + F which becomes 8×F = A + 5×F Subtract 5×F from both sides of the above equation: 8×F – 5×F = A + 5×F – 5×F which makes 3×F = A


  

Hint #7


Substitute 3×F for A in eq.2b: D + 3×F = 3×F Subtract 3×F from both sides of the above equation: D + 3×F – 3×F = 3×F – 3×F which makes D = 0


  

Solution

Substitute 3×F for A, 9×F for B, 2×F for C, 0 for D, and 4×F for E in eq.1: 3×F + 9×F + 2×F + 0 + 4×F + F = 19 which simplifies to 19×F = 19 Divide both sides of the equation above by 19: 19×F ÷ 19 = 19 ÷ 19 which means F = 1 making A = 3×F = 3 × 1 = 3 B = 9×F = 9 × 1 = 9 C = 2×F = 2 × 1 = 2 E = 4×F = 4 × 1 = 4 and ABCDEF = 392041