Puzzle for March 30, 2021  ( )

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Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A + B + C + D + E + F = 35 eq.2) D = A + F eq.3) E = A + B eq.4) C + F = B + D eq.5) D + F = B + E – D eq.6) B + D + F = C + E – B

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

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Hint #1


In eq.6, replace B + D with C + F (from eq.4): C + F + F = C + E – B which becomes C + 2×F = C + E – B In the equation above, subtract C from both sides, and add B to both sides: C + 2×F + B – C = C + E – B + B – C which becomes eq.6a) 2×F + B = E


  

Hint #2


In eq.3, replace E with 2×F + B (from eq.6a): 2×F + B = A + B Subtract B from both sides of the above equation: 2×F + B – B = A + B – B which makes 2×F = A


  

Hint #3


In eq.2, substitute 2×F for A: D = 2×F + F which makes D = 3×F


  

Hint #4


Substitute 3×F for D, and 2×F + B for E (from eq.6a) in eq.5: 3×F + F = B + 2×F + B – 3×F which becomes 4×F = 2×B – F Add F to both sides of the above equation: 4×F + F = 2×B – F + F which becomes 5×F = 2×B Divide both sides by 2: 5×F ÷ 2 = 2×B ÷ 2 which makes 2½×F = B


  

Hint #5


Substitute 2½×F for B in eq.6a: 2×F + 2½×F = E which makes 4½×F = E


  

Hint #6


Substitute 2½×F for B, and 3×F for D in eq.4: C + F = 2½×F + 3×F which becomes C + F = 5½×F Subtract F from each side of the equation above: C + F – F = 5½×F – F which makes C = 4½×F


  

Solution

Substitute 2×F for A, 2½×F for B, 4½×F for C and E, and 3×F for D in eq.1: 2×F + 2½×F + 4½×F + 3×F + 4½×F + F = 35 which simplifies to 17½×F = 35 Divide both sides of the equation above by 17½: 17½×F ÷ 17½ = 35 ÷ 17½ which means F = 2 making A = 2×F = 2 × 2 = 4 B = 2½×F = 2½ × 2 = 5 C = E = 4½×F = 4½ × 2 = 9 D = 3×F = 3 × 2 = 6 and ABCDEF = 459692