Puzzle for May 1, 2019  ( )

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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) A + B + C = D + E + F eq.3) C + D = F eq.4) B + D = A + E eq.5) E + F = A + C eq.6)* A + B + C = CD

A, B, C, D, E, and F each represent a one-digit non-negative integer.
*  CD is a 2-digit number (not C×D).

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Hint #1


In eq.2, replace E + F with A + C (from eq.5): A + B + C = D + A + C Subtract both A and C from each side of the above equation: A + B + C - A - C = D + A + C - A - C which makes B = D


  

Hint #2


In eq.3, replace D with B: C + B = F which is the same as B + C = F In eq.2, replace B + C with F: A + F = D + E + F Subtract F from both sides of the above equation: A + F - F = D + E + F - F which becomes eq.2a) A = D + E


  

Hint #3


In eq.4, substitute D + E for A (from eq.2a): B + D = D + E + E Subtract D from each side: B + D - D = D + E + E - D which makes B = 2×E which also makes D = B = 2×E


  

Hint #4


Substitute 2×E for B and D in eq.4: 2×E + 2×E = A + E Subtract E from each side of the above equation: 2×E + 2×E - E = A + E - E which makes 3×E = A


  

Hint #5


eq.6 may be written as: A + B + C = 10×C + D Subtract C from each side of the equation above: A + B + C - C = 10×C + D - C which becomes eq.6a) A + B = 9×C + D


  

Hint #6


Substitute 3×E for A, and 2×E for B and D in eq.6a: 3×E + 2×E = 9×C + 2×E Subtract 2×E from each side of the above equation: 3×E + 2×E - 2×E = 9×C + 2×E - 2×E which simplifies to 3×E = 9×C Divide both sides by 9: 3×E ÷ 9 = 9×C ÷ 9 which means ⅓×E = C


  

Hint #7


Substitute ⅓×E for C, and 2×E for D in eq.3: ⅓×E + 2×E = F which means 2⅓×E = F


  

Solution

Substitute 3×E for A, 2×E for B and D, ⅓×E for C, and 2⅓×E for F in eq.1: 3×E + 2×E + ⅓×E + 2×E + E + 2⅓×E = 32 which becomes 10⅔×E = 32 Divide both sides by 10⅔: 10⅔×E ÷ 10⅔ = 32 ÷ 10⅔ which means E = 3 making A = 3×E = 3 × 3 = 9 B = D = 2×E = 2 × 3 = 6 C = ⅓×E = ⅓ × 3 = 1 F = 2⅓×E = 2⅓ × 3 = 7 and ABCDEF = 961637