Puzzle for October 20, 2021  ( )

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

eq.1) A + B + C + D + E + F = 30 eq.2) C + F = A + B eq.3) E – A = A + D eq.4) B + F = A + E eq.5) A + C + D = B + E eq.6) D = average (B, E, F)

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


  

Hint #2


Subtract A from both sides of eq.3: E – A – A = A + D – A which becomes eq.3a) E – 2×A = D   In eq.6a, substitute (E – 2×A) for D (from eq.3a), and A + E for B + F (from eq.4): 3×(E – 2×A) = A + E + E which becomes 3×E – 6×A = A + 2×E In the above equation, add 6×A to both sides, and subtract 2×E from both sides: 3×E – 6×A + 6×A – 2×E = A + 2×E + 6×A – 2×E which makes E = 7×A


  

Hint #3


In eq.3a, replace E with 7×A: 7×A – 2×A = D which makes 5×A = D


  

Hint #4


In eq.5, replace D with 5×A, and E with 7×A: A + C + 5×A = B + 7×A which becomes C + 6×A = B + 7×A Subtract 6×A from each side of the equation above: C + 6×A – 6×A = B + 7×A – 6×A which becomes C = B + A which may be written as eq.5a) C = A + B


  

Hint #5


In eq.2, substitute C for A + B (from eq.5a): C + F = C Subtract C from both sides of the above equation: C + F – C = C – C which makes F = 0


  

Hint #6


Substitute 0 for F, and 7×A for E in eq.4: B + 0 = A + 7×A which becomes B = 8×A


  

Hint #7


Substitute 8×A for B into eq.5a: C = A + 8×A which makes C = 9×A


  

Solution

Substitute 8×A for B, 9×A for C, 5×A for D, 7×A for E, and 0 for F in eq.1: A + 8×A + 9×A + 5×A + 7×A + 0 = 30 which simplifies to 30×A = 30 Divide both sides of the above equation by 30: 30×A ÷ 30 = 30 ÷ 30 which means A = 1 making B = 8×A = 8 × 1 = 8 C = 9×A = 9 × 1 = 9 D = 5×A = 5 × 1 = 5 E = 7×A = 7 × 1 = 7 and ABCDEF = 189570