Puzzle for August 16, 2022  ( )

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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) D = E + F eq.3) C = A + D eq.4) B = C + F eq.5) B + F = C + E eq.6) D + E = B + 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.5, replace B with C + F (from eq.4): C + F + F = C + E which becomes C + 2×F = C + E Subtract C from each side of the equation above: C + 2×F – C = C + E – C which makes 2×F = E


  

Hint #2


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


  

Hint #3


In eq.6, substitute 3×F for D, C + F for B (from eq.4), and 2×F for E: 3×F + 2×F = C + F + C – 3×F which becomes 5×F = 2×C – 2×F Add 2×F to both sides of the above equation: 5×F + 2×F = 2×C – 2×F + 2×F which makes 7×F = 2×C Divide both sides by 2: 7×F ÷ 2 = 2×C ÷ 2 which makes 3½×F = C


  

Hint #4


Substitute 3½×F for C, and 3×F for D in eq.3: 3½×F = A + 3×F Subtract 3×F from each side of the above equation: 3½×F – 3×F = A + 3×F – 3×F which makes ½×F = A


  

Hint #5


Substitute 3½×F for C in eq.4: B = 3½×F + F which makes B = 4½×F


  

Solution

Substitute ½×F for A, 4½×F for B, 3½×F for C, 3×F for D, and 2×F for E in eq.1: ½×F + 4½×F + 3½×F + 3×F + 2×F + F = 29 which simplifies to 14½×F = 29 Divide both sides of the above equation by 14½: 14½×F ÷ 14½ = 29 ÷ 14½ which means F = 2 making A = ½×F = ½ × 2 = 1 B = 4½×F = 4½ × 2 = 9 C = 3½×F = 3½ × 2 = 7 D = 3×F = 3 × 2 = 6 E = 2×F = 2 × 2 = 4 and ABCDEF = 197642