Puzzle for February 17, 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 = F eq.3) A + C = B + F eq.4) E + F = C + D eq.5) C + F = B + D + E 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.3, replace F with B + C (from eq.2): A + C = B + B + C which becomes A + C = 2×B + C Subtract C from both sides of the above equation: A + C – C = 2×B + C – C which makes A = 2×B


  

Hint #2


In eq.6, replace A with 2×B: D + E = 2×B + B + C which becomes eq.6a) D + E = 3×B + C


  

Hint #3


In eq.5, substitute 3×B + C for D + E (from eq.6a): C + F = B + 3×B + C which becomes C + F = 4×B + C Subtract C from each side of the equation above: C + F – C = 4×B + C – C which makes F = 4×B


  

Hint #4


Substitute 2×B for A, and 4×B for F in eq.3: 2×B + C = B + 4×B which becomes 2×B + C = 5×B Subtract 2×B from each side of the above equation: 2×B + C – 2×B = 5×B – 2×B which makes C = 3×B


  

Hint #5


Substitute 4×B for F, and 3×B for C in eq.4: E + 4×B = 3×B + D Subtract 3×B from both sides of the equation above: E + 4×B – 3×B = 3×B + D – 3×B which becomes eq.4a) E + B = D


  

Hint #6


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


  

Hint #7


Substitute 2½×B for E in eq.4a: 2½×B + B = D which makes 3½×B = D


  

Solution

Substitute 2×B for A, 3×B for C, 3½×B for D, 2½×B for E, and 4×B for F in eq.1: 2×B + B + 3×B + 3½×B + 2½×B + 4×B = 32 which simplifies to 16×B = 32 Divide both sides of the above equation by 16: 16×B ÷ 16 = 32 ÷ 16 which means B = 2 making A = 2×B = 2 × 2 = 4 C = 3×B = 3 × 2 = 6 D = 3½×B = 3½ × 2 = 7 E = 2½×B = 2½ × 2 = 5 F = 4×B = 4 × 2 = 8 and ABCDEF = 426758