Puzzle for July 12, 2023  ( )

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

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

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Hint #1


Subtract the left and right sides of eq.4 from the left and right sides of eq.5, respectively: B + D + F - (B + D - F) = A + C + E - (A + E + F) which becomes B + D + F - B - D + F = A + C + E - A - E - F which becomes 2×F = C - F Add F to both sides of the above equation: 2×F + F = C - F + F which makes 3×F = C


  

Hint #2


In eq.6, replace C with 3×F: A = 3×F ÷ F which makes A = 3


  

Hint #3


eq.5 may be re-written as: B + D + F = A + E + C In the above equation, replace A + E with D + F (from eq.2): B + D + F = D + F + C Subtract D and F from each side: B + D + F - D - F = D + F + C - D - F which simplifies to B = C and makes B = C = 3×F


  

Hint #4


In eq.3, substitute 3 for A, and 3×F for B: E + F = 3 + 3×F Subtract F from both sides of the equation above: E + F - F = 3 + 3×F - F which becomes eq.3a) E = 3 + 2×F


  

Hint #5


Substitute 3×F for B, 3 for A, and 3 + 2×F for E (from eq.3a) in eq.4: 3×F + D - F = 3 + 3 + 2×F + F which becomes 2×F + D = 6 + 3×F Subtract 2×F from both sides of the above equation: 2×F + D - 2×F = 6 + 3×F - 2×F which becomes eq.4a) D = 6 + F


  

Hint #6


Substitute 3 for A, 3×F for B and C, 6 + F for D (from eq.4a), and 3 + 2×F for E (from eq.3a) in eq.1: 3 + 3×F + 3×F + 6 + F + 3 + 2×F + F = 32 which simplifies to 12 + 10×F = 32 Subtract 12 from each side of the above equation: 12 + 10×F - 12 = 32 - 12 which makes 10×F = 20 Divide both sides by 10: 10×F ÷ 10 = 20 ÷ 10 which means F = 2


  

Solution

Since F = 2, then: B = C = 3×F = 3 × 2 = 6 D = 6 + F = 6 + 2 = 8 (from eq.4a) E = 3 + 2×F = 3 + 2×2 = 3 + 4 = 7 (from eq.3a) and ABCDEF = 366872