Puzzle for July 20, 2022 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit non-negative integer.
Scratchpad
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Hint #1
Add A to both sides of eq.4: B – A + A = D + F – C + A which becomes B = D + F – C + A which may be written as eq.4a) B = A + F + D – C
Hint #2
In eq.4a, replace A + F with B + C + D (from eq.5): B = B + C + D + D – C which becomes B = B + 2×D Subtract B from each side of the equation above: B – B = B + 2×D – B which makes 0 = 2×D which means 0 = D
Hint #3
In eq.2, replace D with 0: 0 + E = A + C which becomes eq.2a) E = A + C
Hint #4
In eq.3, substitute A + C for E (from eq.2a), and 0 for D: A + A + C = 0 + F which becomes eq.3a) 2×A + C = F
Hint #5
In eq.4a, substitute 2×A + C for F (from eq.3a), and 0 for D: B = A + 2×A + C + 0 – C which makes B = 3×A
Hint #6
Substitute A + C for E (from eq.2a), 2×A + C for F (from eq.3a), 3×A for B, and 0 for D in eq.6: A + C + 2×A + C – 3×A = 3×A + C – 0 which becomes 2×C = 3×A + C Subtract C from each side of the above equation: 2×C – C = 3×A + C – C which makes C = 3×A
Hint #7
Substitute 3×A for C in eq.3a: 2×A + 3×A = F which makes 5×A = F
Hint #8
Substitute 3×A for C in eq.2a: E = A + 3×A which makes E = 4×A
Solution
Substitute 3×A for B and C, 0 for D, 4×A for E, and 5×A for F in eq.0: A + 3×A + 3×A + 0 + 4×A + 5×A = 16 which simplifies to 16×A = 16 Divide both sides of the above equation by 16: 16×A ÷ 16 = 16 ÷ 16 which means A = 1 making B = C = 3×A = 3 × 1 = 3 E = 4×A = 4 × 1 = 4 F = 5×A = 5 × 1 = 5 and ABCDEF = 133045