Puzzle for July 9, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 41 eq.2) C + F = D eq.3) A + C = B + E – A eq.4) B + D – F = C + E eq.5) D – C + F = C – A – F eq.6)* DE + F = D + EF

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

Scratchpad

 

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Hint #1


eq.6 may be written as: 10×D + E + F = D + 10×E + F Subtract D, E, and F from both sides of the above equation: 10×D + E + F – D – E – F = D + 10×E + F – D – E – F which simplifies to 9×D = 9×E Divide both sides by 9: 9×D ÷ 9 = 9×E ÷ 9 which makes D = E


  

Hint #2


In eq.4, replace E with D: B + D – F = C + D Subtract D from each side of the equation above: B + D – F – D = C + D – D which becomes eq.4a) B – F = C


  

Hint #3


In eq.2, replace C with B – F (from eq.4a): B – F + F = D which becomes B = D


  

Hint #4


In eq.3, substitute D for B and E: A + C = D + D – A which becomes A + C = 2×D – A Add A to both sides of the equation above: A + C + A = 2×D – A + A which becomes eq.3a) 2×A + C = 2×D   Divide both sides of eq.3a by 2: (2×A + C) ÷ 2 = 2×D ÷ 2 which becomes eq.3b) A + ½×C = D


  

Hint #5


Substitute (C + F) for D (from eq.2) in eq.3a: 2×A + C = 2×(C + F) which is equivalent to 2×A + C = 2×C + 2×F Subtract 2×C from each side of the above equation: 2×A + C – 2×C = 2×C + 2×F – 2×C which becomes 2×A – C = 2×F Divide both sides by 2: (2×A – C) ÷ 2 = 2×F ÷ 2 which becomes eq.3c) A – ½×C = F


  

Hint #6


Substitute A + ½×C for D (from eq.3b), and (A – ½×C) for F (from eq.3c) in eq.5: A + ½×C – C + (A – ½×C) = C – A – (A – ½×C) which becomes 2×A – C = C – A – A + ½×C which becomes 2×A – C = 1½×C – 2×A Add C and 2×A to both sides of the equation above: 2×A – C + C + 2×A = 1½×C – 2×A + C + 2×A which becomes 4×A = 2½×C Divide both sides by 4: 4×A ÷ 4 = 2½×C ÷ 4 which makes eq.5a) A = ⅝×C


  

Hint #7


Substitute ⅝×C for A in eq.3c: ⅝×C – ½×C = F which makes ⅛×C = F Multiply both sides of the equation above by 8: 8 × ⅛×C = 8 × F which makes C = 8×F


  

Hint #8


Substitute (8×F) for C in eq.5a: A = ⅝×(8×F) which makes A = 5×F


  

Hint #9


Substitute 5×F for A, and (8×F) for C in eq.3b: 5×F + ½×(8×F) = D which is equivalent to 5×F + 4×F = D which makes 9×F = D and also makes E = D = B = 9×F


  

Solution

Substitute 5×F for A, 9×F for B and D and E, and 8×F for C in eq.1: 5×F + 9×F + 8×F + 9×F + 9×F + F = 41 which simplifies to 41×F = 41 Divide each side of the equation above by 41: 41×F ÷ 41 = 41 ÷ 41 which means F = 1 making A = 5×F = 5 × 1 = 5 B = D = E = 9×F = 9 × 1 = 9 C = 8×F = 8 × 1 = 8 and ABCDEF = 598991