Puzzle for May 6, 2019  ( )

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

eq.1) A + B + C + D + E + F = 26 eq.2) F = A + C + D eq.3) A + B = D + F eq.4) C + D + E = B eq.5) B - A = A + F eq.6) D = A × C

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

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Hint #1


Add A to both sides of eq.5: B - A + A = A + F + A which becomes eq.5a) B = 2×A + F In eq.3, replace B with 2×A + F: A + 2×A + F = D + F which becomes 3×A + F = D + F Subtract F from both sides of the above equation: 3×A + F - F = D + F - F which makes 3×A = D


  

Hint #2


In eq.6, replace D with 3×A: 3×A = C × A Divide both sides of the above equation by A: 3×A ÷ A = C × A ÷ A which makes 3 = C


  

Hint #3


Substitute 3 for C, and 3×A for D in eq.2: F = 3 + A + 3×A which means eq.2a) F = 3 + 4×A


  

Hint #4


Substitute 3 + 4×A for F (from eq.2a) in eq.5a: B = 2×A + 3 + 4×A which means eq.5b) B = 3 + 6×A


  

Hint #5


Substitute 3 for C, 3×A for D, and 3 + 6×A for B (from eq.5b) in eq.4: 3 + 3×A + E = 3 + 6×A Subtract both 3 and 3×A from each side of the above equation: 3 + 3×A + E - 3 - 3×A = 3 + 6×A - 3 - 3×A which makes E = 3×A


  

Solution

Substitute 3 + 6×A for B (from eq.5b), 3 for C, 3×A for D and E, and 3 + 4×A for F (from eq.2a) in eq.1: A + 3 + 6×A + 3 + 3×A + 3×A + 3 + 4×A = 26 which becomes 17×A + 9 = 26 Subtract 9 from both sides of the equation above: 17×A + 9 - 9 = 26 - 9 which simplifies to 17×A = 17 Divide both sides by 17: 17×A ÷ 17 = 17 ÷ 17 which means A = 1 making B = 3 + 6×A = 3 + 6×1 = 3 + 6 = 9 (from eq.5b) D = E = 3×A = 3×1 = 3 F = 3 + 4×A = 3 + 4×1 = 3 + 4 = 7 (from eq.2a) and ABCDEF = 193337