Puzzle for November 4, 2021  ( )

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

eq.1) A + B + C + D + E + F = 19 eq.2) C = D + E eq.3) E = D + F eq.4) B + F = A + D eq.5) C + D = A + F eq.6) A – B + C = B + E + F

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

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Hint #1


In eq.2, replace E with D + F (from eq.3): C = D + D + F which becomes eq.2a) C = 2×D + F


  

Hint #2


In eq.5, replace C with 2×D + F (from eq.2a): 2×D + F + D = A + F which becomes 3×D + F = A + F Subtract F from each side of the above equation: 3×D + F – F = A + F – F which makes 3×D = A


  

Hint #3


In eq.4, substitute 3×D for A: B + F = 3×D + D which becomes eq.4a) B + F = 4×D


  

Hint #4


eq.6 may be written as: A – B + C = E + B + F Substitute 3×D for A, and 4×D for B + F (from eq.4a) in the above equation: 3×D – B + C = E + 4×D Subtract 3×D from each side: 3×D – B + C – 3×D = E + 4×D – 3×D which becomes –B + C = E + D which may be written as eq.6a) C – B = D + E


  

Hint #5


Substitute C for D + E (from eq.2) into eq.6a: C – B = C Subtract C from each side of the above equation: C – B – C = C – C which makes –B = 0 which means B = 0


  

Hint #6


Substitute 0 for B in eq.4a: 0 + F = 4×D which makes F = 4×D


  

Hint #7


Substitute 4×D for F in eq.2a: C = 2×D + 4×D which makes C = 6×D


  

Hint #8


Substitute 4×D for F in eq.3: E = D + 4×D which makes E = 5×D


  

Solution

Substitute 3×D for A, 0 for B, 6×D for C, 5×D for E, and 4×D for F in eq.1: 3×D + 0 + 6×D + D + 5×D + 4×D = 19 which simplifies to 19×D = 19 Divide both sides of the above equation by 19: 19×D ÷ 19 = 19 ÷ 19 which means D = 1 making A = 3×D = 3 × 1 = 3 C = 6×D = 6 × 1 = 6 E = 5×D = 5 × 1 = 5 F = 4×D = 4 × 1 = 4 and ABCDEF = 306154