Puzzle for January 15, 2021  ( )

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

eq.1) A + B + C + D + E + F = 38 eq.2) B + F = A + D eq.3) E + F = C eq.4) D – E = B – C eq.5) C = D + (A ÷ F) eq.6) B – A = average (A, C, F)

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

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Hint #1


In eq.4, substitute (E + F) for C (from eq.3): D – E = B – (E + F) which is equivalent to D – E = B – E – F Add E to both sides of the above equation: D – E + E = B – E – F + E which becomes eq.4a) D = B – F


  

Hint #2


In eq.2, replace D with B – F (from eq.4a): B + F = A + B – F In the above equation, subtract B from both sides, and add F to both sides: B + F – B + F = A + B – F – B + F which makes 2×F = A


  

Hint #3


In eq.5, replace A with 2×F: C = D + (2×F ÷ F) which becomes eq.5a) C = D + 2


  

Hint #4


eq.6 may be written as: B – A = (A + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × (B – A) = 3 × (A + C + F) ÷ 3 which becomes eq.6a) 3×B – 3×A = A + C + F


  

Hint #5


In eq.6a, substitute (2×F) for A, and D + 2 for C (from eq.5a): 3×B – 3×(2×F) = (2×F) + D + 2 + F which becomes 3×B – 6×F = 3×F + D + 2 Subtract 3×F from each side of the equation above: 3×B – 6×F – 3×F = 3×F + D + 2 – 3×F which becomes eq.6b) 3×B – 9×F = D + 2


  

Hint #6


Substitute B – F for D (from eq.4a) in eq.6b: 3×B – 9×F = B – F + 2 In the equation above, add 9×F to both sides, and subtract B from both sides: 3×B – 9×F + 9×F – B = B – F + 2 + 9×F – B which becomes 2×B = 2 + 8×F Divide both sides by 2: 2×B ÷ 2 = (2 + 8×F) ÷ 2 which becomes eq.6c) B = 1 + 4×F


  

Hint #7


Substitute 1 + 4×F for B (from eq.6c) in eq.4a: D = 1 + 4×F – F which becomes eq.4b) D = 1 + 3×F


  

Hint #8


Substitute 1 + 3×F for D (from eq.4b) in eq.5a: C = 1 + 3×F + 2 which becomes eq.5b) C = 3 + 3×F


  

Hint #9


Substitute 3 + 3×F for C (from eq.5b) in eq.3: E + F = 3 + 3×F Subtract F from each side of the equation above: E + F – F = 3 + 3×F – F which becomes eq.3a) E = 3 + 2×F


  

Solution

Substitute 2×F for A, 1 + 4×F for B (from eq.6c), 3 + 3×F for C (from eq.5b), 1 + 3×F for D (from eq.4b), and 3 + 2×F for E (from eq.3a) in eq.1: 2×F + 1 + 4×F + 3 + 3×F + 1 + 3×F + 3 + 2×F + F = 38 which simplifies to 15×F + 8 = 38 Subtract 8 from each side of the above equation: 15×F + 8 – 8 = 38 – 8 which makes 15×F = 30 Divide both sides by 15: 15×F ÷ 15 = 30 ÷ 15 which means F = 2 making A = 2×F = 2×2 = 4 B = 1 + 4×F = 1 + 4×2 = 1 + 8 = 9 (from eq.6c) C = 3 + 3×F = 3 + 3×2 = 3 + 6 = 9 (from eq.5b) D = 1 + 3×F = 1 + 3×2 = 1 + 6 = 7 (from eq.4b) E = 3 + 2×F = 3 + 2×2 = 3 + 4 = 7 (from eq.3a) and ABCDEF = 499772