Puzzle for May 9, 2020  ( )

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

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

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

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Hint #1


In eq.3, replace A with C + E (from eq.1): C + E + B + E = C + D which becomes C + B + 2×E = C + D Subtract C from each side of the equation above: C + B + 2×E – C = C + D – C which becomes eq.3a) B + 2×E = D


  

Hint #2


In eq.2, replace D with B + 2×E (from eq.3a): B + 2×E – F = B + F In the equation above, subtract B from each side, and add F to both sides: B + 2×E – F – B + F = B + F – B + F which makes 2×E = 2×F Divide both sides by 2: 2×E ÷ 2 = 2×F ÷ 2 which makes E = F


  

Hint #3


In eq.5, substitute B + 2×E for D (from eq.3a), and E for F: C = B + 2×E + E + E which becomes eq.5a) C = B + 4×E


  

Hint #4


Substitute B + 4×E for C (from eq.5a) in eq.1: B + 4×E + E = A which becomes eq.1a) B + 5×E = A


  

Hint #5


Substitute B + 2×E for D (from eq.3a), B + 5×E for A (from eq.1a), and E for F in eq.4: B + B + 2×E = B + 5×E + E which becomes 2×B + 2×E = B + 6×E Subtract B and 2×E from each side of the equation above: 2×B + 2×E – B – 2×E = B + 6×E – B – 2×E which makes B = 4×E making A = B + 5×E = 4×E + 5×E = 9×E (from eq.1a) C = B + 4×E = 4×E + 4×E = 8×E (from eq.5a) D = B + 2×E = 4×E + 2×E = 6×E (from eq.3a)


  

Solution

eq.6 may be written as: 10×B + C = C × D Substitute (4×E) for B, 8×E for C, and 6×E for D in the above equation: 10×(4×E) + 8×E = 8×E × 6×E which becomes 40×E + 8×E = 48×E×E which makes 48×E = 48×E×E Divide both sides of the equation above by 48×E: 48×E ÷ 48×E = 48×E×E ÷ 48×E which makes 1 = E making A = 9×E = 9 × 1 = 9 B = 4×E = 4 × 1 = 4 C = 8×E = 8 × 1 = 8 D = 6×E = 6 × 1 = 6 F = E = 1 and ABCDEF = 948611