Puzzle for December 6, 2022  ( )

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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) A = C + E eq.3) F = B + C eq.4) D = A + C eq.5) C + D = B + E eq.6) B + F = A + C + E

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

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Hint #1


In eq.4, replace A with C + E (from eq.2): D = C + E + C which becomes eq.4a) D = 2×C + E


  

Hint #2


In eq.5, replace D with 2×C + E (from eq.4a): C + 2×C + E = B + E which becomes 3×C + E = B + E Subtract E from each side of the above equation: 3×C + E - E = B + E - E which makes 3×C = B


  

Hint #3


In eq.3, substitute 3×C for B: F = 3×C + C which makes F = 4×C


  

Hint #4


Substitute 3×C for B, 4×C for F, and A for C + E (from eq.2) in eq.6: 3×C + 4×C = A + A which becomes 7×C = 2×A Divide both sides of the above equation by 2: 7×C ÷ 2 = 2×A ÷ 2 which makes 3½×C = A


  

Hint #5


Substitute 3½×C for A in eq.2: 3½×C = C + E Subtract C from each side of the equation above: 3½×C - C = C + E - C which makes 2½×C = E


  

Hint #6


Substitute 3½×C for A in eq.4: D = 3½×C + C which makes D = 4½×C


  

Solution

Substitute 3½×C for A, 3×C for B, 4½×C for D, 2½×C for E, and 4×C for F in eq.1: 3½×C + 3×C + C + 4½×C + 2½×C + 4×C = 37 which simplifies to 18½×C = 37 Divide both sides of the above equation by 18½: 18½×C ÷ 18½ = 37 ÷ 18½ which means C = 2 making A = 3½×C = 3½ × 2 = 7 B = 3×C = 3 × 2 = 6 D = 4½×C = 4½ × 2 = 9 E = 2½×C = 2½ × 2 = 5 F = 4×C = 4 × 2 = 8 and ABCDEF = 762958