Puzzle for June 14, 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) B = C + D eq.3) E = C + F eq.4) A + D = B + F eq.5) D + F = B – F eq.6) B + D = A + E + F

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

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Hint #1


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


  

Hint #2


In eq.4, replace A with 3×F, and B with C + D (from eq.2): 3×F + D = C + D + F Subtract D and F from each side of the equation above: 3×F + D – D – F = C + D + F – D – F which simplifies to 2×F = C


  

Hint #3


In eq.3, substitute 2×F for C: E = 2×F + F which makes E = 3×F


  

Hint #4


Substitute D + 2×F for B (from eq.5a), and 3×F for A and E in eq.6: D + 2×F + D = 3×F + 3×F + F which becomes 2×D + 2×F = 7×F Subtract 2×F from each side of the equation above: 2×D + 2×F – 2×F = 7×F – 2×F which makes 2×D = 5×F Divide both sides by 2: 2×D ÷ 2 = 5×F ÷ 2 which makes D = 2½×F


  

Hint #5


Substitute 2×F for C, and 2½×F for D in eq.2: B = 2×F + 2½×F which makes B = 4½×F


  

Solution

Substitute 3×F for A and E, 4½×F for B, 2×F for C, and 2½×F for D in eq.1: 3×F + 4½×F + 2×F + 2½×F + 3×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 = E = 3×F = 3 × 2 = 6 B = 4½×F = 4½ × 2 = 9 C = 2×F = 2 × 2 = 4 D = 2½×F = 2½ × 2 = 5 and ABCDEF = 694562