Puzzle for February 5, 2021  ( )

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

eq.1) A + B + C + D + E + F = 37 eq.2) B + C = D + F – B eq.3) E + F = D eq.4) A + C = B + D eq.5) B + C + E = A – C + D eq.6) F = average (A, E)

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 B + C with D + F – B (from eq.2): D + F – B + E = A – C + D Subtract D from each side of the equation above: D + F – B + E – D = A – C + D – D which becomes F – B + E = A – C which may be written as eq.5a) E + F – B = A – C


  

Hint #2


In eq.5a, replace E + F with D (from eq.3): D – B = A – C which may be written as eq.5b) A – C = D – B


  

Hint #3


Add the left and right sides of eq.5b to the left and right sides of eq.4, respectively: A + C + (A – C) = B + D + (D – B) which makes 2×A = 2×D Divide both sides of the above equation by 2: 2×A ÷ 2 = 2×D ÷ 2 which makes A = D


  

Hint #4


In eq.4, substitute A for D: A + C = B + A Subtract A from each side of the above equation: A + C – A = B + A – A which makes C = B


  

Hint #5


eq.6 may be written as: F = (A + E) ÷ 2 Multiply both sides of the equation above by 2: 2 × F = 2 × (A + E) ÷ 2 which becomes 2×F = A + E Subtract E from both sides: 2×F – E = A + E – E which makes 2×F – E = A and also makes eq.6a) D = A = 2×F – E


  

Hint #6


Substitute 2×F – E for D (from eq.6a) in eq.3: E + F = 2×F – E In the above equation, subtract F from each side, and add E to each side: E + F – F + E = 2×F – E – F + E which makes 2×E = F


  

Hint #7


Substitute (2×E) for F in eq.6a: D = A = 2×(2×E) – E which becomes D = A = 4×E – E which makes D = A = 3×E


  

Hint #8


Substitute B for C, 3×E for D, and 2×E for F in eq.2: B + B = 3×E + 2×E – B which becomes 2×B = 5×E – B Add B to each side of the above equation: 2×B + B = 5×E – B + B which makes 3×B = 5×E Divide both sides by 3: 3×B ÷ 3 = 5×E ÷ 3 which makes B = 1⅔×E and also makes C = B = 1⅔×E


  

Solution

Substitute 3×E for A and D, 1⅔×E for B and C, and 2×E for F in eq.1: 3×E + 1⅔×E + 1⅔×E + 3×E + E + 2×E = 37 which simplifies to 12⅓×E = 37 Divide both sides of the equation above by 12⅓: 12⅓×E ÷ 12⅓ = 37 ÷ 12⅓ which means E = 3 making A = D = 3×E = 3 × 3 = 9 B = C = 1⅔×E = 1⅔ × 3 = 5 F = 2×E = 2 × 3 = 6 and ABCDEF = 955936