Puzzle for February 7, 2023  ( )

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

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 C + D with A + E (from eq.3): E + F = A + A + E which becomes E + F = 2×A + E Subtract E from each side of the equation above: E + F - E = 2×A + E - E which makes eq.6a) F = 2×A


  

Hint #2


eq.6 may be written as: E + F = A + D + C In the above equation, replace E + F with B + D (from eq.2), and A + D with B + C (from eq.4): B + D = B + C + C which becomes B + D = B + 2×C Subtract B from each side of the above equation: B + D - B = B + 2×C - B which makes D = 2×C


  

Hint #3


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


  

Hint #4


In eq.3, substitute 2×C for D, and 3×A for E: C + 2×C = A + 3×A which makes 3×C = 4×A Divide both sides of the above equation by 4: 3×C ÷ 4 = 4×A ÷ 4 which makes ¾×C = A


  

Hint #5


Substitute (¾×C) for A in eq.5a: E = 3×(¾×C) which makes E = 2¼×C


  

Hint #6


Substitute (¾×C) for A in eq.6a: F = 2×(¾×C) which makes F = 1½×C


  

Hint #7


Substitute ¾×C for A, and 2×C for D in eq.4: B + C = ¾×C + 2×C which becomes B + C = 2¾×C Subtract C from each side of the equation above: B + C - C = 2¾×C - C which makes B = 1¾×C


  

Solution

Substitute ¾×C for A, 1¾×C for B, 2×C for D, 2¼×C for E, and 1½×C for F in eq.1: ¾×C + 1¾×C + C + 2×C + 2¼×C + 1½×C = 37 which simplifies to 9¼×C = 37 Divide both sides of the above equation by 9¼: 9¼×C ÷ 9¼ = 37 ÷ 9¼ which means C = 4 making A = ¾×C = ¾ × 4 = 3 B = 1¾×C = 1¾ × 4 = 7 D = 2×C = 2 × 4 = 8 E = 2¼×C = 2¼ × 4 = 9 F = 1½×C = 1½ × 4 = 6 and ABCDEF = 374896