Puzzle for November 15, 2021  ( )

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

eq.1) A + B + C + D + E + F = 29 eq.2) C = D + E eq.3) B = A + E eq.4) F = A + C eq.5) A = C + D + E eq.6) A – E = C – D + E

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

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Hint #1


In eq.6, replace A with C + D + E (from eq.5): C + D + E – E = C – D + E which becomes C + D = C – D + E In the above equation, subtract C from both sides, and add D to both sides: C + D – C + D = C – D + E – C + D which becomes 2×D = E


  

Hint #2


In eq.2, replace E with 2×D: C = D + 2×D which makes C = 3×D


  

Hint #3


In eq.5, substitute 3×D for C, and 2×D for E: A = 3×D + D + 2×D which makes A = 6×D


  

Hint #4


Substitute 6×D for A, and 2×D for E in eq.3: B = 6×D + 2×D which makes B = 8×D


  

Hint #5


Substitute 6×D for A, and 3×D for C in eq.4: F = 6×D + 3×D which makes F = 9×D


  

Solution

Substitute 6×D for A, 8×D for B, 3×D for C, 2×D for E, and 9×D for F in eq.1: 6×D + 8×D + 3×D + D + 2×D + 9×D = 29 which simplifies to 29×D = 29 Divide both sides of the above equation by 29: 29×D ÷ 29 = 29 ÷ 29 which means D = 1 making A = 6×D = 6 × 1 = 6 B = 8×D = 8 × 1 = 8 C = 3×D = 3 × 1 = 3 E = 2×D = 2 × 1 = 2 F = 9×D = 9 × 1 = 9 and ABCDEF = 683129