Puzzle for October 1, 2022  ( )

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

eq.1) A + B + C + D + E + F = 22 eq.2) E = C + F eq.3) A - C = B + F eq.4) D - F = B + C - E + F eq.5) E + F - B = A + B + C eq.6) F = average (A, C, D)

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 (C + F) for E (from eq.2): D - F = B + C - (C + F) + F which becomes D - F = B + C - C - F + F which becomes D - F = B Add F to both sides of the above equation: D - F + F = B + F which becomes eq.4a) D = B + F


  

Hint #2


In eq.3, replace B + F with D (from eq.4a): A - C = D Add C to both sides of the above equation: A - C + C = D + C which becomes A = D + C which may be written as eq.3a) A = C + D


  

Hint #3


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


  

Hint #4


In eq.6a, replace C + D with A (from eq.3a): 3×F = A + A which makes 3×F = 2×A Divide both sides of the above equation by 2: 3×F ÷ 2 = 2×A ÷ 2 which makes eq.6b) 1½×F = A


  

Hint #5


In eq.3, substitute 1½×F for A (from eq.6b): 1½×F - C = B + F In the equation above, add C to both sides, and subtract F from both sides: 1½×F - C + C - F = B + F + C - F which becomes eq.3b) ½×F = B + C


  

Hint #6


Substitute 1½×F for A (from eq.6b), and ½×F for B + C (from eq.3b) in eq.5: E + F - B = 1½×F + ½×F which becomes E + F - B = 2×F In the equation above, subtract F from both sides, and add B to both sides: E + F - B - F + B = 2×F - F + B which becomes eq.5a) E = F + B


  

Hint #7


Substitute F + B for E (from eq.5a) into eq.2: F + B = C + F Subtract F from each side of the equation above: F + B - F = C + F - F which makes B = C


  

Hint #8


Substitute B for C in eq.3b: ½×F = B + B which makes ½×F = 2×B Multiply both sides of the above equation by 2: 2 × (½×F) = 2 × (2×B) which makes F = 4×B


  

Hint #9


Substitute (4×B) for F in eq.6b: 1½×(4×B) = A which makes 6×B = A


  

Hint #10


Substitute 4×B for F in eq.5a: E = 4×B + B which makes E = 5×B


  

Hint #11


Substitute 4×B for F in eq.4a: D = B + 4×B which makes D = 5×B


  

Solution

Substitute 6×B for A, B for C, 5×B for D and E, and 4×B for F in eq.1: 6×B + B + B + 5×B + 5×B + 4×B = 22 which simplifies to 22×B = 22 Divide both sides of the above equation by 22: 22×B ÷ 22 = 22 ÷ 22 which means B = 1 making A = 6×B = 6 × 1 = 6 C = B = 1 D = E = 5×B = 5 × 1 = 5 F = 4×B = 4 × 1 = 4 and ABCDEF = 611554