Puzzle for March 26, 2023  ( )

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

eq.1) A + B + C + D + E + F = 34 eq.2) A + E = C - E + F eq.3) C + D = B + E + F eq.4) A + D = C + F - A eq.5) E = average (B, 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 E to both sides of eq.2: A + E + E = C - E + F + E which becomes eq.2a) A + 2×E = C + F   Add A to both sides of eq.4: A + D + A = C + F - A + A which becomes eq.4a) 2×A + D = C + F


  

Hint #2


In eq.4a, replace C + F with A + 2×E (from eq.2a): 2×A + D = A + 2×E Subtract A from each side of the above equation: 2×A + D - A = A + 2×E - A which becomes eq.2b) A + D = 2×E


  

Hint #3


eq.5 may be written as: E = (B + C + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × E = 4 × (B + C + D + F) ÷ 4 which becomes eq.5a) 4×E = B + C + D + F which may be written as eq.5b) 2×(2×E) = B + C + D + F


  

Hint #4


In eq.5b, replace 2×E with A + D (from eq.2b): 2×(A + D) = B + C + D + F which becomes 2×A + 2×D = B + C + D + F Subtract D from both sides of the above equation: 2×A + 2×D - D = B + C + D + F - D which becomes eq.5c) 2×A + D = B + C + F


  

Hint #5


In eq.5c, substitute C + F for 2×A + D (from eq.4a): C + F = B + C + F Subtract both C and F from each side of the above equation: C + F - C - F = B + C + F - C - F which makes 0 = B


  

Hint #6


Substitute 0 for B in eq.3: C + D = 0 + E + F which becomes eq.3a) C + D = E + F


  

Hint #7


Substitute 0 for B, and E + F for C + D (from eq.3a) in eq.5a: 4×E = 0 + E + F + F which becomes 4×E = E + 2×F Subtract E from each side of the equation above: 4×E - E = E + 2×F - E which makes 3×E = 2×F Divide both sides by 2: 3×E ÷ 2 = 2×F ÷ 2 which makes eq.5d) 1½×E = F


  

Hint #8


eq.1 may be written as: A + D + B + C + E + F = 34 In the above equation, substitute 2×E for A + D (from eq.2b), 0 for B, and 1½×E for F: 2×E + 0 + C + E + 1½×E = 34 which becomes 4½×E + C = 34 which may be written as eq.1a) 3×(1½×E) + C = 34


  

Hint #9


Substitute F for (1½×E) in eq.1a: 3×F + C = 34 Subtract 3×F from both sides of the equation above: 3×F + C - 3×F = 34 - 3×F which becomes eq.1b) C = 34 - 3×F


  

Hint #10


To make eq.1b true, check several possible values for F and C:   If F = 9, then C = 34 - 3×F = 34 - 3×9 = 34 - 27 = 7 If F = 8, then C = 34 - 3×F = 34 - 3×8 = 34 - 24 = 10 If F < 8, then C > 10   Since C must be a one-digit non-negative integer, this means C < 10 which makes C = 7 and F = 9


  

Hint #11


Substitute 9 for F in eq.5d: 1½×E = 9 Divide both sides of the above equation by 1½: 1½×E ÷ 1½ = 9 ÷ 1½ which makes E = 6


  

Hint #12


Substitute 7 for C, 6 for E, and 9 for F in eq.3a: 7 + D = 6 + 9 which becomes 7 + D = 15 Subtract 7 from each side of the equation above: 7 + D - 7 = 15 - 7 which makes D = 8


  

Solution

Substitute 8 for D, and 6 for E in eq.2b: A + 8 = 2×6 which becomes A + 8 = 12 Subtract 8 from each side of the above equation: A + 8 - 8 = 12 - 8 which makes A = 4 and makes ABCDEF = 407869