Puzzle for December 2, 2022  ( )

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

eq.1) A + B + C + D + E + F = 23 eq.2) E = A + D eq.3) C - F = D + F eq.4) D - F = A + B - D eq.5) E + F - A = A + D - F eq.6) A = average (B, C, E, F)

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

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Hint #1


In eq.5, replace E with A + D (from eq.2): A + D + F - A = A + D - F which becomes D + F = A + D - F In the above equation, subtract D from both sides, and add F to both sides: D + F - D + F = A + D - F - D + F which makes 2×F = A


  

Hint #2


In eq.2, replace A with 2×F: E = 2×F + D which may be written as eq.2a) E = D + 2×F


  

Hint #3


Add F to both sides of eq.3: C - F + F = D + F + F which becomes C = D + 2×F In the above equation, substitute E for D + 2×F (from eq.2a): C = E


  

Hint #4


In eq.4, replace A with 2×F: D - F = 2×F + B - D In the above equation, subtract 2×F from both sides, and add D to both sides: D - F - 2×F + D = 2×F + B - D - 2×F + D which becomes eq.4a) 2×D - 3×F = B


  

Hint #5


eq.6 may be written as: A = (B + C + E + F) ÷ 4 Mutiply both sides of the above equation by 4: 4 × A = 4 × (B + C + E + F) ÷ 4 which becomes eq.6a) 4×A = B + C + E + F


  

Hint #6


Substitute E for C in eq.6a: 4×A = B + E + E + F which becomes eq.6b) 4×A = B + 2×E + F


  

Hint #7


Substitute (2×F) for A, 2×D - 3×F for B (from eq.4a), and (D + 2×F) for E (from eq.2a) in eq.6b: 4×(2×F) = 2×D - 3×F + 2×(D + 2×F) + F which becomes 8×F = 2×D - 3×F + 2×D + 4×F + F which becomes 8×F = 4×D + 2×F Subtract 2×F from each side of the equation above: 8×F - 2×F = 4×D + 2×F - 2×F which makes 6×F = 4×D Divide both sides by 4: 6×F ÷ 4 = 4×D ÷ 4 which makes 1½×F = D


  

Hint #8


Substitute 1½×F for D in eq.2a: E = 1½×F + 2×F which makes E = 3½×F and also makes C = E = 3½×F


  

Hint #9


Substitute (1½×F) for D in eq.4a: 2×(1½×F) - 3×F = B which becomes 3×F - 3×F = B which makes 0 = B


  

Solution

Substitute 2×F for A, 0 for B, 3½×F for C and E, and 1½×F for D in eq.1: 2×F + 0 + 3½×F + 1½×F + 3½×F + F = 23 which simplifies to 11½×F = 23 Divide both sides of the above equation by 11½: 11½×F ÷ 11½ = 23 ÷ 11½ which means F = 2 making A = 2×F = 2 × 2 = 4 C = E = 3½×F = 3½ × 2 = 7 D = 1½×F = 1½ × 2 = 3 and ABCDEF = 407372