Puzzle for August 18, 2019 ( )
Scratchpad
Find the 6-digit number ABCDEF by solving the following equations:
A, B, C, D, E, and F each represent a one-digit positive integer.
Scratchpad
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Hint #1
eq.5 may be written as: D × E = B + F + C In the above equation, replace B + F with A + D + E (from eq.3): D × E = A + D + E + C which is the same as eq.5a) D × E = A + C + E + D
Hint #2
In eq.5a, replace A + C + E with D × F (from eq.6): D × E = D × F + D which may be written as D × E = D × (F + 1) Divide both sides of the above equation by D: D × E ÷ D = D × (F + 1) ÷ D which makes eq.5b) E = F + 1
Hint #3
In eq.2, substitute F + 1 for E (from eq.5b): D = F + 1 + F which makes eq.2a) D = 2×F + 1
Hint #4
Substitute F + 1 for E (from eq.5b), and 2×F + 1 for D (from eq.2a) in eq.4: C + F + 1 = 2×F + 1 + F Subtract F and 1 from each side of the equation above: C + F + 1 – F – 1 = 2×F + 1 + F – F – 1 which simplifies to C = 2×F
Hint #5
Substitute (2×F + 1) for D (from eq.2a), 2×F for C, and F + 1 for E (from eq.5b) in eq.6: (2×F + 1) × F = A + 2×F + F + 1 which becomes 2×F² + F = A + 3×F + 1 Subtract 3×F and 1 from both sides of the above equation: 2×F² + F – 3×F – 1 = A + 3×F + 1 – 3×F – 1 which becomes eq.6a) 2×F² – 2×F – 1 = A
Hint #6
Substitute 2×F² – 2×F – 1 for A (from eq.6a), 2×F + 1 for D (from eq.2a), and F + 1 for E (from eq.5b) in eq.3: 2×F² – 2×F – 1 + 2×F + 1 + F + 1 = B + F which becomes 2×F² + F + 1 = B + F Subtract F from both sides of the equation above: 2×F² + F + 1 – F = B + F – F which makes eq.5c) 2×F² + 1 = B
Hint #7
Substitute 2×F² – 2×F – 1 for A (from eq.6a), 2×F² + 1 for B (from eq.5c), 2×F for C, 2×F + 1 for D (from eq.2a), and F + 1 for E (from eq.5b) in eq.1: 2×F² – 2×F – 1 + 2×F² + 1 + 2×F + 2×F + 1 + F + 1 + F = 26 which simplifies to 4×F² + 4×F + 2 = 26 Subtract 26 from both sides of the above equation: 4×F² + 4×F + 2 – 26 = 26 – 26 which becomes 4×F² + 4×F – 24 = 0 Divide both sides by 4: (4×F² + 4×F – 24) ÷ 4 = 0 ÷ 4 which makes eq.1a) F² + F – 6 = 0
Solution
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for F in eq.1a yields: F = {(–1)×(1) ± sq.rt.[(1)² – (4 × (1) × (–6))]} ÷ (2 × (1)) which becomes F = {–1 ± sq.rt.[1 – (–24)]} ÷ 2 which becomes F = {–1 ± sq.rt.(25)} ÷ 2 which becomes F = (–1 ± 5) ÷ 2 In the above equation, either F = (–1 + 5) ÷ 2 = 4 ÷ 2 = 2 or F = (–1 – 5) ÷ 2 = –6 ÷ 2 = –3 Since F must be a positive integer, then F ≠ –3 and therefore makes F = 2 making A = 2×F² – 2×F – 1 = 2×2² – 2×2 – 1 = 2×4 – 4 – 1 = 8 – 5 = 3 (from eq.6a) B = 2×F² + 1 = 2×2² + 1 = 2×4 + 1 = 8 + 1 = 9 (from eq.5c) C = 2×F = 2×2 = 4 D = 2×F + 1 = 2×2 + 1 = 4 + 1 = 5 (from eq.2a) E = F + 1 = 2 + 1 = 3 (from eq.5b) and ABCDEF = 394532