Puzzle for August 16, 2020 ( )
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 D from both sides of eq.2: D + E – D = A – D which becomes eq.2a) E = A – D In eq.4, replace E with A – D (from eq.2a): B – D = A + A – D Add D to both sides of the equation above: B – D + D = A + A – D + D which makes eq.4a) B = 2×A
Hint #2
In eq.5, replace B with 2×A, and replace C with A + D (from eq.3): A × D = 2×A + A + D which becomes A × D = 3×A + D (implies D ≠ 3) Subtract 3×A from each side of the equation above: A × D – 3×A = 3×A + D – 3×A which becomes A × D – 3×A = D which may be written as A × (D – 3) = D Divide both sides by (D – 3): A × (D – 3) ÷ (D – 3) = D ÷ (D – 3) which makes eq.5a) A = D ÷ (D – 3)
Hint #3
In eq.2a, substitute D ÷ (D – 3) for A (from eq.5a): eq.2b) E = D ÷ (D – 3) – D
Hint #4
In eq.4a, substitute D ÷ (D – 3) for A (from eq.5a): eq.4b) B = 2×D ÷ (D – 3)
Hint #5
In eq.3, substitute D ÷ (D – 3) for A (from eq.5a): eq.3a) C = D ÷ (D – 3) + D
Hint #6
Substitute 2×A for B, and D ÷ (D – 3) + D for C (from eq.3a) in eq.6: F + (2×A ÷ A) = D ÷ (D – 3) + D + D – F which becomes F + 2 = D ÷ (D – 3) + 2×D – F In the equation above, subtract 2 from each side, and add F to each side: F + 2 – 2 + F = D ÷ (D – 3) + 2×D – F – 2 + F which becomes 2×F = D ÷ (D – 3) + 2×D – 2 Divide both sides by 2: 2×F ÷ 2 = (D ÷ (D – 3) + 2×D – 2) ÷ 2 which becomes eq.6a) F = ½×D ÷ (D – 3) + D – 1
Hint #7
Substitute D ÷ (D – 3) for A (from eq.5a), 2×D ÷ (D – 3) for B (from eq.4b), D ÷ (D – 3) + D for C (from eq.3a), D ÷ (D – 3) – D for E (from eq.3a), ½×D ÷ (D – 3) + D – 1 for F (from eq.6a) in eq.1: D ÷ (D – 3) + 2×D ÷ (D – 3) + D ÷ (D – 3) + D + D + D ÷ (D – 3) – D + ½×D ÷ (D – 3) + D – 1 = 29 which becomes 5½×D ÷ (D – 3) + 2×D – 1 = 29 In the above equation, subtract 2×D from both sides, and add 1 to both sides: 5½×D ÷ (D – 3) + 2×D – 1 – 2×D + 1 = 29 – 2×D + 1 which becomes eq.1a) 5½×D ÷ (D – 3) = 30 – 2×D
Hint #8
Multiply both sides of eq.1a by (D – 3): 5½×D ÷ (D – 3) × (D – 3) = (30 – 2×D) × (D – 3) which becomes 5½×D = 30×D – 90 – 2×D² + 6×D which becomes 5½×D = 36×D – 90 – 2×D² Subtract 5½×D from both sides: 5½×D – 5½×D = 36×D – 90 – 2×D² – 5½×D which becomes eq.1b) 0 = –2×D² + 30½×D – 90
Solution
eq.1b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.1b yields: D = { (–1)×(30½) ± sq.rt.[(30½)² – (4 × (–2) × (–90))] } ÷ (2 × (–2)) which becomes D = {–30½ ± sq.rt.(930¼ – 720)} ÷ (–4) which becomes D = {–30½ ± sq.rt.(210¼)} ÷ (–4) which becomes D = (–30½ ± 14½) ÷ (–4) In the above equation, either D = (–30½ + 14½) ÷ (–4) = –16 ÷ (–4) = 4 or D = (–30½ – 14½) ÷ (–4) = –45 ÷ (–4) = 11¼ Since D must be a one-digit integer, then D ≠ 11¼ and therefore makes D = 4 making A = D ÷ (D – 3) = 4 ÷ (4 – 3) = 4 ÷ (1) = 4 (from eq.5a) B = 2×(D ÷ (D – 3)) = 2×(4 ÷ (4 – 3)) = 2×(4 ÷ (1)) = 2×4 = 8 (from eq.4b) C = D ÷ (D – 3) + D = 4 ÷ (4 – 3) + 4 = 4 ÷ (1) + 4 = 4 + 4 = 8 (from eq.3a) E = D ÷ (D – 3) – D = 4 ÷ (4 – 3) – 4 = 4 ÷ (1) – 4 = 4 – 4 = 0 (from eq.2b) F = ½×D ÷ (D – 3) + D – 1 = ½×4 ÷ (4 – 3) + 4 – 1 = 2 ÷ (1) + 3 = 2 + 3 = 5 (from eq.6a) and ABCDEF = 488405