Puzzle for December 7, 2021  ( )

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

eq.1) A + B + C + D + E + F = 31 eq.2) C + F = E eq.3) D + F = B eq.4) E + F = A + B eq.5) B + D = A + F eq.6) C + D = A + B + F

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


  

Hint #2


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


  

Hint #3


In eq.3, substitute 3×F for D: 3×F + F = B which makes 4×F = B


  

Hint #4


Substitute 4×F for B, and 3×F for D in eq.5: 4×F + 3×F = A + F which becomes 7×F = A + F Subtract F from each side of the above equation: 7×F – F = A + F – F which makes 6×F = A


  

Hint #5


Substitute 6×F for A, and 4×F for B in eq.4: E + F = 6×F + 4×F which becomes E + F = 10×F Subtract F from both sides of the above equation: E + F – F = 10×F – F which makes E = 9×F


  

Hint #6


Substitute 9×F for E in eq.2: C + F = 9×F Subtract F from each side of the equation above: C + F – F = 9×F – F which makes C = 8×F


  

Solution

Substitute 6×F for A, 4×F for B, 8×F for C, 3×F for D, and 9×F for E in eq.1: 6×F + 4×F + 8×F + 3×F + 9×F + F = 31 which simplifies to 31×F = 31 Divide both sides of the above equation by 31: 31×F ÷ 31 = 31 ÷ 31 which means F = 1 making A = 6×F = 6 × 1 = 6 B = 4×F = 4 × 1 = 4 C = 8×F = 8 × 1 = 8 D = 3×F = 3 × 1 = 3 E = 9×F = 9 × 1 = 9 and ABCDEF = 648391