Puzzle for November 14, 2022  ( )

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

eq.1) A + B + C + D + E + F = 32 eq.2) F = B + D eq.3) E = A + C eq.4) A = C + D eq.5) B + D = A + C eq.6) A + D = B + C - D

A, B, C, D, E, and F each represent a one-digit non-negative integer.
 
Many thanks to Abby S (age 12) for sending us this puzzle! Thank you so much, Abby!

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Hint #1


In eq.5, replace A with C + D (from eq.4): B + D = C + D + C which becomes B + D = 2×C + D Subtract D from each side of the above equation: B + D - D = 2×C + D - D which becomes eq.5a) B = 2×C


  

Hint #2


In eq.6, replace A with C + D (from eq.4), and B with 2×C: C + D + D = 2×C + C - D which becomes C + 2×D = 3×C - D In the equation above, subtract C from both sides, and add D to both sides: C + 2×D - C + D = 3×C - D - C + D which becomes 3×D = 2×C Divide both sides by 2: 3×D ÷ 2 = 2×C ÷ 2 which makes 1½×D = C


  

Hint #3


In eq.4, substitute 1½×D for C: A = 1½×D + D which makes A = 2½×D


  

Hint #4


In eq.5a, substitute (1½×D) for C: B = 2×(1½×D) which makes B = 3×D


  

Hint #5


Substitute 3×D for B in eq.2: F = 3×D + D which makes F = 4×D


  

Hint #6


Substitute 2½×D for A, and 1½×D for C in eq.3: E = 2½×D + 1½×D which makes E = 4×D


  

Solution

Substitute 2½×D for A, 3×D for B, 1½×D for C, and 4×D for E and F in eq.1: 2½×D + 3×D + 1½×D + D + 4×D + 4×D = 32 which simplifies to 16×D = 32 Divide both sides of the above equation by 16: 16×D ÷ 16 = 32 ÷ 16 which means D = 2 making A = 2½×D = 2½ × 2 = 5 B = 3×D = 3 × 2 = 6 C = 1½×D = 1½ × 2 = 3 E = F = 4×D = 4 × 2 = 8 and ABCDEF = 563288