Puzzle for March 5, 2022  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) C + D = A + B eq.3) E – F = D – E eq.4) B + E = A + C + F eq.5) C – F = A – D + E eq.6) D = average (A, B, C)

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

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Hint #1


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


  

Hint #2


In eq.6a, replace A + B with C + D (from eq.2): 3×D = C + D + C which becomes 3×D = 2×C + D Subtract D from each side of the equation above: 3×D – D = 2×C + D – D which makes 2×D = 2×C Divide both sides by 2: 2×D ÷ 2 = 2×C ÷ 2 which makes D = C


  

Hint #3


Add F and D to both sides of eq.5: C – F + F + D = A – D + E + F + D which becomes C + D = A + E + F In the above equation, replace C + D with A + B (from eq.2): A + B = A + E + F Subtract A from both sides: A + B – A = A + E + F – A which becomes eq.5a) B = E + F


  

Hint #4


In eq.4, substitute E + F for B (from eq.5a): E + F + E = A + C + F which becomes 2×E + F = A + C + F Subtract F from each side of the equation above: 2×E + F – F = A + C + F – F which becomes eq.4a) 2×E = A + C


  

Hint #5


Add E to both sides of eq.3: E – F + E = D – E + E which becomes 2×E – F = D In the equation above, substitute A + C for 2×E (from eq.4a), and C for D: A + C – F = C Subtract C from both sides, and add F to both sides: A + C – F – C + F = C – C + F which makes A = F


  

Hint #6


Substitute A for F in eq.5a: eq.5b) B = E + A


  

Hint #7


Substitute C for D, and E + A for B (from eq.5b) in eq.2: C + C = A + E + A which becomes 2×C = 2×A + E Divide both sides of the above equation by 2: 2×C ÷ 2 = (2×A + E) ÷ 2 which becomes eq.2a) C = A + ½×E


  

Hint #8


Substitute A + ½×E for C (from eq.2a) in eq.4a: 2×E = A + A + ½×E which becomes 2×E = 2×A + ½×E Subtract ½×E from each side of the above equation: 2×E – ½×E = 2×A + ½×E – ½×E which makes 1½×E = 2×A Divide both sides by 2 1½×E ÷ 2 = 2×A ÷ 2 which makes ¾×E = A and also makes F = A = ¾×E


  

Hint #9


Substitute ¾×E for A in eq.2a: C = ¾×E + ½×E which makes C = 1¼×E and also makes D = C = 1¼×E


  

Hint #10


Substitute ¾×E for A in eq.5b: B = E + ¾×E which makes B = 1¾×E


  

Solution

Substitute ¾×E for A and F, 1¾×E for B, and 1¼×E for C and D in eq.1: ¾×E + 1¾×E + 1¼×E + 1¼×E + E + ¾×E = 27 which simplifies to 6¾×E = 27 Divide both sides of the above equation by 6¾: 6¾×E ÷ 6¾ = 27 ÷ 6¾ which means E = 4 making A = F = ¾×E = ¾ × 4 = 3 B = 1¾×E = 1¾ × 4 = 7 C = D = 1¼×E = 1¼ × 4 = 5 and ABCDEF = 375543