Puzzle for February 10, 2020  ( )

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

eq.1) A + B + C + D + E + F = 22 eq.2) B – C = C + D eq.3) B + C = A + D eq.4) A + C = D eq.5) C + F = B + D eq.6) A + B = E + F

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

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Hint #1


Add C to both sides of eq.2: B – C + C = C + D + C which becomes eq.2a) B = 2×C + D   In eq.3, replace B with 2×C + D (from eq.2a): 2×C + D + C = A + D Subtract D from both sides of the equation above: 2×C + D + C – D = A + D – D which makes 3×C = A


  

Hint #2


In eq.4, replace A with 3×C: 3×C + C = D which makes 4×C = D


  

Hint #3


In eq.2a, substitute 4×C for D: B = 2×C + 4×C which makes B = 6×C


  

Hint #4


Substitute 6×C for B, and 4×C for D in eq.5: C + F = 6×C + 4×C which becomes C + F = 10×C Subtract C from each side of the above equation: C + F – C = 10×C – C which makes F = 9×C


  

Hint #5


Substitute 3×C for A, 6×C for B, and 9×C for F in eq.6: 3×C + 6×C = E + 9×C which becomes 9×C = E + 9×C Subtract 9×C from each side of the above equation: 9×C – 9×C = E + 9×C – 9×C which makes 0 = E


  

Solution

Substitute 3×C for A, 6×C for B, 4×C for D, 0 for E, and 9×C for F in eq.1: 3×C + 6×C + C + 4×C + 0 + 9×C = 22 which simplifies to 22×C = 22 Divide each side of the equation above by 22: 22×C ÷ 22 = 22 ÷ 22 which means C = 1 making A = 3×C = 3 × 1 = 3 B = 6×C = 6 × 1 = 6 D = 4×C = 4 × 1 = 4 F = 9×C = 9 × 1 = 9 and ABCDEF = 361409