Puzzle for June 25, 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) E = A + C eq.3) C + E = D + F eq.4) D – B – F = B + C + F eq.5) E + F – B = C + D – F eq.6) B = average (C, D, F)

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

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Hint #1


Add B to both sides of eq.4: D – B – F + B = B + C + F + B which becomes eq.4a) D – F = 2×B + C + F   Subtract C from each side of eq.5: E + F – B – C = C + D – F – C which becomes eq.5a) E + F – B – C = D – F


  

Hint #2


In eq.4a, replace D – F with E + F – B – C (from eq.5a): E + F – B – C = 2×B + C + F In the above equation, subtract F from both sides, and add B and C to both sides: E + F – B – C – F + B + C = 2×B + C + F – F + B + C which simplifies to eq.4b) E = 3×B + 2×C


  

Hint #3


eq.6 may be written as: B = (C + D + F) ÷ 3 Multiply both sides of the above equation by 3: B × 3 = ((C + D + F) ÷ 3) × 3 which becomes eq.6a) 3×B = C + D + F   In the equation above, replace D + F with C + E (from eq.3): 3×B = C + C + E which becomes eq.6b) 3×B = 2×C + E


  

Hint #4


In eq.4b, substitute 2×C + E for 3×B (from eq.6b): E = 2×C + E + 2×C which becomes E = E + 4×C Subtract E from both sides of the equation above: E – E = E + 4×C – E which makes 0 = 4×C which means 0 = C


  

Hint #5


Substitute 0 for C in eq.2: E = A + 0 which makes E = A


  

Hint #6


Substitute 0 for C in eq.4b: E = 3×B + 2×0 which makes E = 3×B and also makes A = E = 3×B


  

Hint #7


Substitute 0 for C in eq.6a: 3×B = 0 + D + F which becomes 3×B = D + F Subtract F from both sides of the above equation: 3×B – F = D + F – F which becomes eq.6c) 3×B – F = D


  

Hint #8


In eq.4a, substitute 3×B – F for D (from eq.6c), and 0 for C: 3×B – F – F = 2×B + 0 + F which becomes 3×B – 2×F = 2×B + F In the above equation, add 2×F to both sides, and subtract 2×B from both sides: 3×B – 2×F + 2×F – 2×B = 2×B + F + 2×F – 2×B which makes B = 3×F and makes A = E = 3×B = 3×(3×F) = 9×F


  

Hint #9


Substitute (3×F) for B in eq.6c: 3×(3×F) – F = D which becomes 9×F – F = D which makes 8×F = D


  

Solution

Substitute 9×F for A and E, 3×F for B, 0 for C, and 8×F for D in eq.1: 9×F + 3×F + 0 + 8×F + 9×F + F = 30 which simplifies to 30×F = 30 Divide both sides of the equation above by 30: 30×F ÷ 30 = 30 ÷ 30 which means F = 1 making A = E = 9×F = 9 × 1 = 9 B = 3×F = 3 × 1 = 3 D = 8×F = 8 × 1 = 8 and ABCDEF = 930891