Puzzle for May 16, 2020  ( )

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

eq.1) A + B + C + D + E + F = 33 eq.2) C + D = E eq.3) E – F = A – D eq.4) B – F = A – B + D eq.5) A – F = C + D – E + F eq.6)* BC – AB = B – A

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

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Hint #1


Add E and F to both sides of eq.5: A – F + E + F = C + D – E + F + E + F which becomes A + E = C + D + 2×F In the above equation, replace C + D with E (from eq.2): A + E = E + 2×F Subtract E from both sides: A + E – E = E + 2×F – E which makes A = 2×F


  

Hint #2


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


  

Hint #3


In eq.4, substitute 2×F for A: B – F = 2×F – B + D Add F and B to both sides of the above equation: B – F + F + B = 2×F – B + D + F + B which becomes eq.4a) 2×B = 3×F + D


  

Hint #4


Substitute C + 2×D for 3×F (from eq.3a) in eq.4a: 2×B = C + 2×D + D which becomes eq.4b) 2×B = C + 3×D


  

Hint #5


eq.6 may be written as: 10×B + C – (10×A + B) = B – A which is the same as 10×B + C – 10×A – B = B – A which becomes 9×B + C – 10×A = B – A In the above equation, add 10×A to both sides, and subtract B from both sides: 9×B + C – 10×A + 10×A – B = B – A + 10×A – B which becomes 8×B + C = 9×A Substitute (2×F) for A: 8×B + C = 9×(2×F) which may be written as eq.6a) 4×(2×B) + C = 6×(3×F)


  

Hint #6


Substitute C + 3×D for 2×B (from eq.4b), and C + 2×D for 3×F (from eq.3a) in eq.6a: 4×(C + 3×D) + C = 6×(C + 2×D) which becomes 4×C + 12×D + C = 6×C + 12×D which becomes 5×C + 12×D = 6×C + 12×D Subtract 5×C and 12×D from both sides of the equation above: 5×C + 12×D – 5×C – 12×D = 6×C + 12×D – 5×C – 12×D which simplifies to 0 = C


  

Hint #7


Substitute 0 for C in eq.3a: 0 + 2×D = 3×F which makes 2×D = 3×F Divide both sides of the above equation by 2: 2×D ÷ 2 = 3×F ÷ 2 which makes D = 1½×F


  

Hint #8


Substitute 1½×F for D in eq.4a: 2×B = 3×F + 1½×F which becomes 2×B = 4½×F Divide both sides of the above equation by 2: 2×B ÷ 2 = 4½×F ÷ 2 which makes B = 2¼×F


  

Hint #9


Substitute 0 for C, and 1½×F for D in eq.2: 0 + 1½×F = E which means 1½×F = E


  

Solution

Substitute 2×F for A, 2¼×F for B, 0 for C, and 1½×F for D and E in eq.1: 2×F + 2¼×F + 0 + 1½×F + 1½×F + F = 33 which simplifies to 8¼×F = 33 Divide both sides of the above equation by 8¼: 8¼×F ÷ 8¼ = 33 ÷ 8¼ which means F = 4 making A = 2×F = 2 × 4 = 8 B = 2¼×F = 2¼ × 4 = 9 D = E = 1½×F = 1½ × 4 = 6 and ABCDEF = 890664