Puzzle for June 28, 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) A + F = D eq.3) B + F = A + D eq.4) C + F = B + D eq.5) D + F = E eq.6) E + F = C

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

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Hint #1


In eq.4, replace C with E + F (from eq.6): E + F + F = B + D which becomes eq.4a) E + 2×F = B + D


  

Hint #2


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


  

Hint #3


In eq.3, substitute 3×F for B, and A + F for D (from eq.2): 3×F + F = A + A + F which becomes 4×F = 2×A + F Subtract F from each side of the equation above: 4×F – F = 2×A + F – F which becomes 3×F = 2×A Divide both sides by 2: 3×F ÷ 2 = 2×A ÷ 2 which makes 1½×F = A


  

Hint #4


Substitute 1½×F for A in eq.2: 1½×F + F = D which makes 2½×F = D


  

Hint #5


Substitute 2½×F for D in eq.5: 2½×F + F = E which makes 3½×F = E


  

Hint #6


Substitute 3½×F for E in eq.6: 3½×F + F = C which makes 4½×F = C


  

Solution

Substitute 1½×F for A, 3×F for B, 4½×F for C, 2½×F for D, and 3½×F for E in eq.1: 1½×F + 3×F + 4½×F + 2½×F + 3½×F + F = 32 which simplifies to 16×F = 32 Divide both sides of the above equation by 16: 16×F ÷ 16 = 32 ÷ 16 which means F = 2 making A = 1½×F = 1½ × 2 = 3 B = 3×F = 3 × 2 = 6 C = 4½×F = 4½ × 2 = 9 D = 2½×F = 2½ × 2 = 5 E = 3½×F = 3½ × 2 = 7 and ABCDEF = 369572