Puzzle for January 10, 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) E = C + D eq.3) A = C + F eq.4) B + D = A + C eq.5) D – F = A – D eq.6) A + F = B + C + D

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

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Hint #1


eq.6 may be written as: A + F = B + D + C In the above equation, replace B + D with A + C (from eq.4): A + F = A + C + C which becomes A + F = A + 2×C Subtract A from each side of the equation above: A + F – A = A + 2×C – A which makes F = 2×C


  

Hint #2


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


  

Hint #3


In eq.5, substitute 2×C for F, and 3×C for A: D – 2×C = 3×C – D Add 2×C and D to both sides of the above equation: D – 2×C + 2×C + D = 3×C – D + 2×C + D which becomes 2×D = 5×C Divide both sides by 2: 2×D ÷ 2 = 5×C ÷ 2 which makes D = 2½×C


  

Hint #4


Substitute 2½×C for D in eq.2: E = C + 2½×C which makes E = 3½×C


  

Hint #5


Substitute 2½×C for D, and 3×C for A in eq.4: B + 2½×C = 3×C + C which becomes B + 2½×C = 4×C Subtract 2½×C from each side of the equation above: B + 2½×C – 2½×C = 4×C – 2½×C which makes B = 1½×C


  

Solution

Substitute 3×C for A, 1½×C for B, 2½×C for D, 3½×C for E, and 2×C for F in eq.1: 3×C + 1½×C + C + 2½×C + 3½×C + 2×C = 27 which simplifies to 13½×C = 27 Divide both sides of the above equation by 13½: 13½×C ÷ 13½ = 27 ÷ 13½ which means C = 2 making A = 3×C = 3 × 2 = 6 B = 1½×C = 1½ × 2 = 3 D = 2½×C = 2½ × 2 = 5 E = 3½×C = 3½ × 2 = 7 F = 2×C = 2 × 2 = 4 and ABCDEF = 632574