Puzzle for February 17, 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) A + B = C + F eq.3) B + C + F = D eq.4) E + F = A + C - F eq.5) A + D - E = C + E eq.6)* DE - C + F = AB + C

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  AB and DE are 2-digit numbers (not A×B or D×E).

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Hint #1


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


  

Hint #2


Subtract A from each side of eq.4: E + F - A = A + C - F - A which becomes E + F - A = C - F Add the left and right sides of eq.2 to the left and right sides of the above equation, respectively: E + F - A + A + B = C - F + C + F which simplifies to eq.4a) E + F + B = 2×C


  

Hint #3


eq.6 may be written as: 10×D + E - C + F = 10×A + B + C Add C to each side of the above equation: 10×D + E - C + F + C = 10×A + B + C + C which becomes 10×D + E + F = 10×A + B + 2×C In the above equation, replace 2×C with E + F + B (from eq.4a): 10×D + E + F = 10×A + B + E + F + B which becomes 10×D + E + F = 10×A + 2×B + E + F Subtract both E and F from each side: 10×D + E + F - E - F = 10×A + 2×B + E + F - E - F which simplifies to eq.6a) 10×D = 10×A + 2×B


  

Hint #4


Substitute (A + 2×B) for D (from eq.3a) in eq.6a: 10×(A + 2×B) = 10×A + 2×B which is equivalent to 10×A + 20×B = 10×A + 2×B Subtract 10×A and 2×B from each side of the above equation: 10×A + 20×B - 10×A - 2×B = 10×A + 2×B - 10×A - 2×B which simplifies to 18×B = 0 which means B = 0


  

Hint #5


In eq.3a, replace B with 0: A + 2×0 = D which makes A = D


  

Hint #6


Add E to both sides of eq.3: B + C + F + E = D + E which may be written as C + E + F + B = D + E In the above equation, substitute 2×C for E + F + B (from eq.4a): C + 2×C = D + E Subtract D from each side: C + 2×C - D = D + E - D which becomes eq.3b) 3×C - D = E


  

Hint #7


Substitute D for A, and (3×C - D) for E (from eq.3b) in eq.5: D + D - (3×C - D) = C + 3×C - D which becomes 2×D - 3×C + D = 4×C - D which becomes 3×D - 3×C = 4×C - D Add 3×C and D to both sides of the above equation: 3×D - 3×C + 3×C + D = 4×C - D + 3×C + D which simplifies to 4×D = 7×C Divide both sides by 4: 4×D ÷ 4 = 7×C ÷ 4 which makes D = 1¾×C


  

Hint #8


Substitute 1¾×C for D in eq.3b: 3×C - 1¾×C = E which makes 1¼×C = E


  

Hint #9


Substitute 0 for B, and 1¾×C for D in eq.3: 0 + C + F = 1¾×C Subtract C from each side: 0 + C + F - C = 1¾×C - C which makes F = ¾×C


  

Hint #10


Substitute 0 for B, and ¾×C for F in eq.2: A + 0 = C + ¾×C which means A = 1¾×C


  

Solution

Substitute 1¾×C for A and D, 0 for B, 1¼×C for E, and ¾×C for F in eq.1: 1¾×C + 0 + C + 1¾×C + 1¼×C + ¾×C = 26 which simplifies to 6½×C = 26 Divide both sides by 6½: 6½×C ÷ 6½ = 26 ÷ 6½ which makes C = 4 making A = D = 1¾×C = 1¾ × 4 = 7 E = 1¼×C = 1¼ × 4 = 5 F = ¾×C = ¾ × 4 = 3 and ABCDEF = 704753