Puzzle for August 20, 2021 ( )
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
Subtract C from each side of eq.3: C + E – C = (C × F) – C which becomes eq.3a) E = (C × F) – C In eq.4, substitute ((C × F) – C) for E (from eq.3a): F – A = ((C × F) – C) ÷ C which becomes F – A = F – 1 Subtract F from each side of the above equation: F – A – F = F – 1 – F which makes –A = –1 which means A = 1
Hint #2
In eq.1, replace A with 1: C = 1 + D Subtract 1 from each side of the above equation: C – 1 = 1 + D – 1 which makes eq.1a) C – 1 = D
Hint #3
In eq.5, replace A with 1: 1 + C = B ÷ C Multiply both sides of the above equation by C: (1 + C) × C = (B ÷ C) × C which becomes eq.5a) C + C² = B
Hint #4
In eq.2, substitute C + C² for B (from eq.5a): E = C + C² + C which becomes eq.2a) E = C² + 2×C
Hint #5
Substitute C² + 2×C for E (from eq.2a) in eq.3: C + C² + 2×C = C × F which becomes 3×C + C² = C × F Since C ≠ 0, divide both sides of the above equation by C: (3×C + C²) ÷ C = (C × F) ÷ C which makes eq.3a) 3 + C = F
Hint #6
In eq.6, substitute (C + C²) for B (from eq.5a), (3 + C) for F (from eq.3a), C – 1 for D (from eq.1a), and C² + 2×C for E (from eq.2a): ((C + C²) × (3 + C)) ÷ C = (C + C²) + C – 1 + C² + 2×C which becomes (C × (1 + C) × (3 + C)) ÷ C = 2×C² + 4×C – 1 which becomes (1 + C) × (3 + C) = 2×C² + 4×C – 1 which becomes 3 + C + 3×C + C² = 2×C² + 4×C – 1 which becomes 3 + 4×C + C² = 2×C² + 4×C – 1 In the above equation, subtract 4×C and C² from each side, and add 1 to each side: 3 + 4×C + C² – 4×C – C² + 1 = 2×C² + 4×C – 1 – 4×C – C² + 1 which simplifies to 4 = C² which means C = ± 2
Solution
Since C is a non-negative integer, then: C ≠ –2 which means C = 2 making B = C + C² = 2 + 2² = 2 + 4 = 6 (from eq.5a) D = C – 1 = 2 – 1 = 1 (from eq.1a) E = C² + 2×C = 2² + 2×2 = 4 + 4 = 8 (from eq.2a) F = 3 + C = 3 + 2 = 5 (from eq.3a) and ABCDEEF = 162185