Puzzle for July 23, 2020  ( )

Scratchpad

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

eq.1) A + B + C + D + E + F = 32 eq.2) A + D = C + F eq.3) B + C = E eq.4) D – E + F = A + B + C eq.5) C + D – A = A + F eq.6) E + F = A + C + D – E

A, B, C, D, E, and F each represent a one-digit non-negative integer.

Scratchpad

 

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Hint #1


eq.6 may be re-written as: E + F = A + D + C – E In the equation above, replace A + D with C + F (from eq.2): E + F = C + F + C – E which becomes E + F = 2×C + F – E Subtract F from both sides, and add E to both sides: E + F – F + E = 2×C + F – E – F + E which simplifies to 2×E = 2×C Divide both sides by 2: 2×E ÷ 2 = 2×C ÷ 2 which makes E = C


  

Hint #2


In eq.3, replace E with C: B + C = C Subtract C from both sides of the equation above: B + C – C = C – C which makes B = 0


  

Hint #3


Add A to both sides of eq.5: C + D – A + A = A + F + A which becomes C + D = 2×A + F In eq.6, substitute 2×A + F for C + D: E + F = A + 2×A + F – E which becomes E + F = 3×A + F – E In the above equation, subtract F from both sides, and add E to both sides: E + F – F + E = 3×A + F – E – F + E which makes 2×E = 3×A Divide both sides by 2: 2×E ÷ 2 = 3×A ÷ 2 which makes E = 1½×A and also makes C = E = 1½×A


  

Hint #4


Substitute 1½×A for C in eq.2: A + D = 1½×A + F Subtract A from both sides of the above equation: A + D – A = 1½×A + F – A which becomes eq.2a) D = ½×A + F


  

Hint #5


Substitute 1½×A for both E and C, and 0 for B in eq.4: D – 1½×A + F = A + 0 + 1½×A which becomes D – 1½×A + F = 2½×A Add 1½×A to each side of the equation above: D – 1½×A + F + 1½×A = 2½×A + 1½×A which becomes eq.4a) D + F = 4×A


  

Hint #6


Substitute ½×A + F for D (from eq.2a) in eq.4a: ½×A + F + F = 4×A which becomes ½×A + 2×F = 4×A Subtract ½×A from each side of the above equation: ½×A + 2×F – ½×A = 4×A – ½×A which makes 2×F = 3½×A Divide both sides by 2: 2×F ÷ 2 = 3½×A ÷ 2 which makes F = 1¾×A


  

Hint #7


Substitute 1¾×A for F in eq.2a: D = ½×A + 1¾×A which makes D = 2¼×A


  

Solution

Substitute 0 for B, 1½×A for C and E, 2¼×A for D, and 1¾×A for F in eq.1: A + 0 + 1½×A + 2¼×A + 1½×A + 1¾×A = 32 which simplifies to 8×A = 32 Divide both sides of the above equation by 8: 8×A ÷ 8 = 32 ÷ 8 which means A = 4 making C = E = 1½×A = 1½ × 4 = 6 D = 2¼×A = 2¼ × 4 = 9 F = 1¾×A = 1¾ × 4 = 7 and ABCDEF = 406967