Puzzle for August 11, 2021  ( )

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

eq.1) A + B + C + D + E + F = 30 eq.2) B + E = C eq.3) E + F = D eq.4) A = B + F eq.5) C + D = A + F eq.6) D + E = A + B + C

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 D with E + F (from eq.3), A with B + F (from eq.4), and C with B + E (from eq.2): E + F + E = B + F + B + B + E which becomes 2×E + F = 3×B + F + E Subtract F and E from both sides of the above equation: 2×E + F – F – E = 3×B + F + E – F – E which makes E = 3×B


  

Hint #2


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


  

Hint #3


In eq.5, substitute E + F for D (from eq.3): C + E + F = A + F Subtract F from each side of the equation above: C + E + F – F = A + F – F which becomes eq.5a) C + E = A


  

Hint #4


Substitute 4×B for C, and 3×B for E in eq.5a: 4×B + 3×B = A which makes 7×B = A


  

Hint #5


Substitute 7×B for A in eq.4: 7×B = B + F Subtract B from both sides of the above equation: 7×B – B = B + F – B which makes 6×B = F


  

Hint #6


Substitute 3×B for E, and 6×B for F in eq.3: 3×B + 6×B = D which makes 9×B = D


  

Solution

Substitute 7×B for A, 4×B for C, 9×B for D, 3×B for E, and 6×B for F in eq.1: 7×B + B + 4×B + 9×B + 3×B + 6×B = 30 which simplifies to 30×B = 30 Divide both sides of the equation above by 30: 30×B ÷ 30 = 30 ÷ 30 which means B = 1 making A = 7×B = 7 × 1 = 7 C = 4×B = 4 × 1 = 4 D = 9×B = 9 × 1 = 9 E = 3×B = 3 × 1 = 3 F = 6×B = 6 × 1 = 6 and ABCDEF = 714936