Puzzle for June 20, 2022  ( )

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

eq.1) A + B + C + D + E + F = 27 eq.2) A = E + F eq.3) B = C + E eq.4) C = A + E eq.5) E = D – B – E eq.6) D – B = F – E

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

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Hint #1


In eq.5, replace D – B with F – E (from eq.6): E = F – E – E which becomes E = F – 2×E Add 2×E to both sides of the above equation: E + 2×E = F – 2×E + 2×E which becomes 3×E = F


  

Hint #2


In eq.2, replace F with 3×E: A = E + 3×E which makes A = 4×E


  

Hint #3


In eq.4, substitute 4×E for A: C = 4×E + E which makes C = 5×E


  

Hint #4


Substitute 5×E for C in eq.3: B = 5×E + E which makes B = 6×E


  

Hint #5


Substitute 6×E for B in eq.5: E = D – 6×E – E which becomes E = D – 7×E Add 7×E to both sides of the above equation: E + 7×E = D – 7×E + 7×E which makes 8×E = D


  

Solution

Substitute 4×E for A, 6×E for B, 5×E for C, 8×E for D, and 3×E for F in eq.1: 4×E + 6×E + 5×E + 8×E + E + 3×E = 27 which simplifies to 27×E = 27 Divide both sides of the above equation by 27: 27×E ÷ 27 = 27 ÷ 27 which means E = 1 making A = 4×E = 4 × 1 = 4 B = 6×E = 6 × 1 = 6 C = 5×E = 5 × 1 = 5 D = 8×E = 8 × 1 = 8 F = 3×E = 3 × 1 = 3 and ABCDEF = 465813