Puzzle for July 16, 2019  ( )

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

eq.1) A + B + C + D + E + F = 20 eq.2) C = D - E eq.3) A = B + F eq.4) D = A + C eq.5)* DE = AB + F eq.6) B - F = D ÷ E

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.2, replace D with A + C (from eq.4): C = A + C - E In the above equation, add E to both sides, and subtract C from each side: C + E - C = A + C - E + E - C which makes E = A


  

Hint #2


eq.5 may be written as: 10×D + E = 10×A + B + F In the above equation, replace E with A, and replace B + F with A (from eq.3): 10×D + A = 10×A + A Subtract A from both sides: 10×D + A - A = 10×A + A - A which becomes 10×D = 10×A Divide both sides by 10: 10×D ÷ 10 = 10×A ÷ 10 which means D = A


  

Hint #3


Substitute A for both D and E in eq.2: C = A - A which means C = 0


  

Hint #4


Substitute E for D in eq.6: B - F = E ÷ E which becomes B - F = 1 Add F to both sides of the above equation: B - F + F = 1 + F which makes eq.6a) B = 1 + F


  

Hint #5


Substitute 1 + F for B in eq.3: A = 1 + F + F which becomes A = 1 + 2×F and which also makes eq.3a) D = E = A = 1 + 2×F


  

Solution

Substitute 1 + 2×F for A and D and E (from eq.3a), 1 + F for B (from eq.6a), and 0 for C in eq.1: 1 + 2×F + 1 + F + 0 + 1 + 2×F + 1 + 2×F + F = 20 which simplifies to 8×F + 4 = 20 Subtract 4 from both sides of the above equation: 8×F + 4 - 4 = 20 - 4 which makes 8×F = 16 Divide both sides by 8: 8×F ÷ 8 = 16 ÷ 8 which means F = 2 making A = D = E = 1 + 2×F = 1 + 2×2 = 5 (from eq.3a) B = 1 + F = 1 + 2 = 3 (from eq.6a) and ABCDEF = 530552