Puzzle for January 20, 2023  ( )

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

eq.1) A + B + C + D + E + F = 45 eq.2) F - A = D - F eq.3) C - E = B - C eq.4) A - E = C - F eq.5) D + F = A + B - D eq.6) C = average (A, E, F)

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

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Hint #1


Add E and F to both sides of eq.4: A - E + E + F = C - F + E + F which becomes eq.4a) A + F = C + E


  

Hint #2


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


  

Hint #3


In eq.6a, replace A + F with C + E (from eq.4a): 3×C = C + E + E which becomes 3×C = C + 2×E Subtract C from both sides of the above equation: 3×C - C = C + 2×E - C which makes 2×C = 2×E Divide both sides by 2: 2×C ÷ 2 = 2×E ÷ 2 which makes C = E


  

Hint #4


In eq.3, replace E with C: C - C = B - C which becomes 0 = B - C Add C to both sides of the equation above: 0 + C = B - C + C which makes C = B


  

Hint #5


Add D to both sides of eq.5: D + F + D = A + B - D + D which becomes eq.5a) 2×D + F = A + B   Add F to both sides of eq.2: F - A + F = D - F + F which becomes eq.2a) 2×F - A = D


  

Hint #6


In eq.5a, substitute (2×F - A) for D (from eq.2a), and C for B: 2×(2×F - A) + F = A + C which becomes 4×F - 2×A + F = A + C which becomes 5×F - 2×A = A + C Add 2×A to both sides of the above equation: 5×F - 2×A + 2×A = A + C + 2×A which becomes eq.5b) 5×F = 3×A + C


  

Hint #7


Substitute C for E in eq.4: A - C = C - F Add C to both sides of the equation above: A - C + C = C - F + C which becomes eq.4b) A = 2×C - F


  

Hint #8


Substitute (2×C - F) for A (from eq.4b) in eq.5b: 5×F = 3×(2×C - F) + C which becomes 5×F = 6×C - 3×F + C which becomes 5×F = 7×C - 3×F Add 3×F to both sides of the above equation: 5×F + 3×F = 7×C - 3×F + 3×F which becomes 8×F = 7×C Divide both sides by 8: 8×F ÷ 8 = 7×C ÷ 8 which makes ⅞×C = F


  

Hint #9


Substitute ⅞×C for F in eq.4b: A = 2×C - ⅞×C which makes A = 1⅛×C


  

Hint #10


Substitute (⅞×C) for F, and 1⅛×C for A in eq.2a: 2×(⅞×C) - 1⅛×C = D which becomes 1¾×C - 1⅛×C = D which makes ⅝×C = D


  

Solution

Substitute 1⅛×C for A, C for B and E, ⅝×C for D, and ⅞×C for F in eq.1: 1⅛×C + C + C + ⅝×C + C + ⅞×C = 45 which simplifies to 5⅝×C = 45 Divide both sides of the above equation by 5⅝: 5⅝×C ÷ 5⅝ = 45 ÷ 5⅝ which means C = 8 making A = 1⅛×C = 1⅛ × 8 = 9 B = E = C = 8 D = ⅝×C = ⅝ × 8 = 5 F = ⅞×C = ⅞ × 8 = 7 and ABCDEF = 988587