Puzzle for October 6, 2021  ( )

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

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

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Hint #1


In eq.6, replace B with A + E (from eq.3): A + E – C = A – E In the above equation, subtract A from each side, and add E and C to both sides: A + E – C – A + E + C = A – E – A + E + C which simplifies to eq.6a) 2×E = C


  

Hint #2


In eq.2, replace C with 2×E (from eq.6a): 2×E + E = D which makes eq.2a) 3×E = D


  

Hint #3


In eq.4, substitute C + E for D (from eq.2): C + E + F = A + C Subtract C from each side of the above equation: C + E + F – C = A + C – C which becomes eq.4a) E + F = A


  

Hint #4


Substitute E + F for A (from eq.4a), and 2×E for C (from eq.6a) in eq.5: E + F + F = 2×E – F which becomes E + 2×F = 2×E – F In the equation above, subtract E from both sides, and add F to both sides: E + 2×F – E + F = 2×E – F – E + F which makes 3×F = E


  

Hint #5


Substitute (3×F) for E in eq.6a: 2×(3×F) = C which makes 6×F = C


  

Hint #6


Substitute (3×F) for E in eq.2a: 3×(3×F) = D which makes 9×F = D


  

Hint #7


Substitute 3×F for E in eq.4a: 3×F + F = A which makes 4×F = A


  

Hint #8


Substitute 4×F for A, and 3×F for E in eq.3: 4×F + 3×F = B which makes 7×F = B


  

Solution

Substitute 4×F for A, 7×F for B, 6×F for C, 9×F for D, and 3×F for E in eq.1: 4×F + 7×F + 6×F + 9×F + 3×F + F = 30 which simplifies to 30×F = 30 Divide both sides of the above equation by 30: 30×F ÷ 30 = 30 ÷ 30 which means F = 1 making A = 4×F = 4 × 1 = 4 B = 7×F = 7 × 1 = 7 C = 6×F = 6 × 1 = 6 D = 9×F = 9 × 1 = 9 E = 3×F = 3 × 1 = 3 and ABCDEF = 476931