Puzzle for July 3, 2021  ( )

Scratchpad

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

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

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

Scratchpad

 

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Hint #1


Add B, D, and E to both sides of eq.1: A – B + C – D + B + D + E = D – E + B + D + E which simplifies to A + C + E = 2×D + B In eq.4, replace A + C + E with 2×D + B: D + F – C = 2×D + B Subtract D from both sides of the above equation: D + F – C – D = 2×D + B – D which becomes F – C = D + B which may be written as eq.4a) F – C = B + D


  

Hint #2


In eq.2, replace B + D with F – C (from eq.4a): F + C = F – C In the equation above, subtract F from each side, and add C to each side: F + C – F + C = F – C – F + C which makes 2×C = 0 which means C = 0


  

Hint #3


In eq.4a, replace C with 0: F – 0 = B + D which becomes eq.4b) F = B + D


  

Hint #4


In eq.3, substitute 0 for C, and B + D for F (from eq.4b): 0 + D + E = B + B + D which becomes D + E = 2×B + D Subtract D from both sides of the equation above: D + E – D = 2×B + D – D which makes eq.3a) E = 2×B


  

Hint #5


Substitute B + D for F (from eq.4b), 0 for C, and 2×B for E in eq.4: D + B + D – 0 = A + 0 + 2×B which becomes B + 2×D = A + 2×B Subtract B from each side of the equation above: B + 2×D – B = A + 2×B – B which becomes eq.4c) 2×D = A + B


  

Hint #6


Substitute 2×D for A + B (from eq.4c), and 0 for C in eq.5: E = (2×D) ÷ (0 + D) which becomes E = 2×D ÷ D which makes E = 2


  

Hint #7


Substitute 2 for E in eq.3a: 2 = 2×B Divide both sides of the equation above by 2: 2 ÷ 2 = 2×B ÷ 2 which makes 1 = B


  

Hint #8


Substitute 2 for E, and 1 for B in eq.6: 2 ÷ 1 = (A ÷ 2) – (D ÷ 2) which becomes 2 = (A – D) ÷ 2 Multiply both sides of the equation above by 2: 2 × 2 = (A – D) ÷ 2 × 2 which becomes 4 = A – D Add D to both sides: 4 + D = A – D + D which becomes eq.6a) 4 + D = A


  

Hint #9


Substitute 4 + D for A (from eq.6a), 1 for B, 0 for C, and 2 for E in eq.1: 4 + D – 1 + 0 – D = D – 2 which becomes 3 = D – 2 Add 2 to both sides of the above equation: 3 + 2 = D – 2 + 2 which makes 5 = D


  

Hint #10


Substitute 5 for D in eq.6a: 4 + 5 = A which makes 9 = A


  

Solution

Substitute 1 for B, and 5 for D in eq.4b: F = 1 + 5 which makes F = 6 and makes ABCDEF = 910526