Puzzle for September 22, 2021  ( )

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

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

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Hint #1


In eq.3, replace D with E + F (from eq.2): E + F + F = B which becomes eq.3a) E + 2×F = B


  

Hint #2


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


  

Hint #3


In eq.6, substitute B + C for D + E (from eq.5), and 3×F for A: C + B + C = 3×F + B which becomes 2×C + B = 3×F + B Subtract B from each side of the above equation: 2×C + B – B = 3×F + B – B which makes 2×C = 3×F Divide both sides by 2: 2×C ÷ 2 = 3×F ÷ 2 which makes C = 1½×F


  

Hint #4


Substitute 3×F for A in eq.4: 3×F + E = B + F Subtract F from each side of the equation above: 3×F + E – F = B + F – F which becomes eq.4a) 2×F + E = B


  

Hint #5


Substitute 2×F + E for B (from eq.4a), and 1½×F for C in eq.5: 2×F + E + 1½×F = D + E which becomes 3½×F + E = D + E Subtract E from each side of the equation above: 3½×F + E – E = D + E – E which makes 3½×F = D


  

Hint #6


Substitute 3½×F for D in eq.3: 3½×F + F = B which makes 4½×F = B


  

Hint #7


Substitute 3½×F for D in eq.2: E + F = 3½×F Subtract F from both sides of the above equation: E + F – F = 3½×F – F which makes E = 2½×F


  

Solution

Substitute 3×F for A, 4½×F for B, 1½×F for C, 3½×F for D, and 2½×F for E in eq.1: 3×F + 4½×F + 1½×F + 3½×F + 2½×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 = 3×F = 3 × 2 = 6 B = 4½×F = 4½ × 2 = 9 C = 1½×F = 1½ × 2 = 3 D = 3½×F = 3½ × 2 = 7 E = 2½×F = 2½ × 2 = 5 and ABCDEF = 693752