Puzzle for March 28, 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
Add B to both sides of eq.2: B – D + B = C – B + B which becomes eq.2a) 2×B – D = C Add D to both sides of eq.2a: 2×B – D + D = C + D which becomes eq.2b) 2×B = C + D
Hint #2
In eq.5, replace C + D with 2×B (from eq.2b): B + F = A + 2×B Subtract B from both sides of the equation above: B + F – B = A + 2×B – B which becomes eq.5a) F = A + B
Hint #3
In eq.4, replace F with A + B (from eq.5a): A + C + A + B = B + E which becomes 2×A + C + B = B + E Subtract B from both sides of the above equation: 2×A + C + B – B = B + E – B which becomes eq.4a) 2×A + C = E
Hint #4
In eq.6, substitute 2×A + C for E (from eq.4a): eq.6a) 2×A + C = A × C Subtract 2×A from both sides of eq.6a: 2×A + C – 2×A = A × C – 2×A which becomes C = A × C – 2×A which may be written as C = A × (C – 2) Divide both sides of the equation above by (C – 2): C ÷ (C – 2) = A × (C – 2) ÷ (C – 2) which becomes eq.6b) C ÷ (C – 2) = A (assumes C ≠ 2)
Hint #5
Check: C ≠ 2 ... Substituting 2 for C in eq.6a would yield: 2×A + 2 = A × 2 which may be written as 2×A + 2 = 2×A Subtracting 2×A from both sides of the equation above would yield: 2×A + 2 – 2×A = 2×A – 2×A which would become 2 = 0 Since 2 ≠ 0, then: C ≠ 2
Hint #6
Substitute C ÷ (C – 2) for A (from eq.6b) in eq.6: E = C ÷ (C – 2) × C which becomes eq.6c) E = C² ÷ (C – 2)
Hint #7
C may be expressed as: C = C × 1 Substitute (C – 2) ÷ (C – 2) for 1 in the equation above: C = C × (C – 2) ÷ (C – 2) which becomes eq.C) C = (C² – 2×C) ÷ (C – 2)
Hint #8
Substitute (C² – 2×C) ÷ (C – 2) for C (from eq.C), and C² ÷ (C – 2) for E (from eq.6c) in eq.3: (C² – 2×C) ÷ (C – 2) + C² ÷ (C – 2) = B + D which becomes (2×C² – 2×C) ÷ (C – 2) = B + D Subtract D from each side of the above equation: (2×C² – 2×C) ÷ (C – 2) – D = B + D – D which becomes eq.3a) (2×C² – 2×C) ÷ (C – 2) – D = B
Hint #9
Substitute ((2×C² – 2×C) ÷ (C – 2) – D) for B (from eq.3a), and (C² – 2×C) ÷ (C – 2) for C (from eq.C) into eq.2b: 2×((2×C² – 2×C) ÷ (C – 2) – D) = (C² – 2×C) ÷ (C – 2) + D which becomes (4×C² – 4×C) ÷ (C – 2) – 2×D = (C² – 2×C) ÷ (C – 2) + D In the equation above, add 2×D to both sides, and subtract ((C² – 2×C) ÷ (C – 2)) from both sides: (4×C² – 4×C) ÷ (C – 2) – 2×D + 2×D – ((C² – 2×C) ÷ (C – 2)) = (C² – 2×C) ÷ (C – 2) + D + 2×D – ((C² – 2×C) ÷ (C – 2)) which becomes ((4×C² – 4×C) – (C² – 2×C)) ÷ (C – 2) = 3×D which becomes (3×C² – 2×C) ÷ (C – 2) = 3×D Divide both sides by 3: ((3×C² – 2×C) ÷ (C – 2)) ÷ 3 = 3×D ÷ 3 which becomes eq.2c) (C² – ⅔×C) ÷ (C – 2) = D
Hint #10
Substitute (C² – ⅔×C) ÷ (C – 2) for D (from eq.2c) into eq.3a: (2×C² – 2×C) ÷ (C – 2) – (C² – ⅔×C) ÷ (C – 2) = B which becomes eq.3b) (C² – 1⅓×C) ÷ (C – 2) = B
Hint #11
Substitute C ÷ (C – 2) for A (from eq.6b), and (C² – 1⅓×C) ÷ (C – 2) for B (from eq.3b) in eq.5a: F = C ÷ (C – 2) + (C² – 1⅓×C) ÷ (C – 2) which becomes eq.5b) F = (C² – ⅓×C) ÷ (C – 2)
Hint #12
In eq.1, substitute -- C ÷ (C – 2) for A (from eq.6b), (C² – 1⅓×C) ÷ (C – 2) for B (from eq.3b), (C² – 2×C) ÷ (C – 2) for C (from eq.C), (C² – ⅔×C) ÷ (C – 2) for D (from eq.2c), C² ÷ (C – 2) for E (from eq.6c), and (C² – ⅓×C) ÷ (C – 2) for F (from eq.5b): C ÷ (C – 2) + (C² – 1⅓×C) ÷ (C – 2) + (C² – 2×C) ÷ (C – 2) + (C² – ⅔×C) ÷ (C – 2) + C² ÷ (C – 2) + (C² – ⅓×C) ÷ (C – 2) = 35 which becomes (C + C² – 1⅓×C + C² – 2×C + C² – ⅔×C + C² + C² – ⅓×C) ÷ (C – 2) = 35 which becomes eq.1a) (5×C² – 3⅓×C) ÷ (C – 2) = 35
Hint #13
Multiply both sides of eq.1a by (C – 2): (5×C² – 3⅓×C) ÷ (C – 2) × (C – 2) = 35 × (C – 2) which becomes 5×C² – 3⅓×C = 35×C – 70 In the equation above, subtract 35×C from both sides, and add 70 to both sides: 5×C² – 3⅓×C – 35×C + 70 = 35×C – 70 – 35×C + 70 which becomes 5×C² – 38⅓×C + 70 = 0 Divide both sides by 5: (5×C² – 38⅓×C + 70) ÷ 5 = 0 ÷ 5 which becomes eq.1b) C² – 7⅔×C + 14 = 0
Solution
eq.1b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.1b yields: C = { (–1)×(–7⅔) ± sq.rt.[(–7⅔)² – (4 × 1 × 14)] } ÷ (2 × (1)) which becomes C = {7⅔ ± sq.rt.((529/9) – 56)} ÷ 2 which may be written as C = {7⅔ ± sq.rt.((529/9) – (504/9))} ÷ 2 which becomes C = {7⅔ ± sq.rt.(25/9)} ÷ 2 which becomes C = (7⅔ ± 1⅔) ÷ 2 In the above equation, either C = (7⅔ + 1⅔) ÷ 2 = 9⅓ ÷ 2 = 4⅔ or C = (7⅔ – 1⅔) ÷ 2 = 6 ÷ 2 = 3 Since C must be an integer, then C ≠ 4⅔ and therefore means C = 3 making A = C ÷ (C – 2) = 3 ÷ (3 – 2) = 3 ÷ 1 = 3 (from eq.6b) B = (C² – 1⅓×C) ÷ (C – 2) = (3² – 1⅓×3) ÷ (3 – 2) = (9 – 4) ÷ 1 = 5 (from eq.3b) D = (C² – ⅔×C) ÷ (C – 2) = (3² – ⅔×3) ÷ (3 – 2) = (9 – 2) ÷ 1 = 7 (from eq.2c) E = C² ÷ (C – 2) = 3² ÷ (3 – 2) = 9 ÷ 1 = 9 (from eq.6c) F = (C² – ⅓×C) ÷ (C – 2) = (3² – ⅓×3) ÷ (3 – 2) = (9 – 1) ÷ 1 = 8 (from eq.5b) and ABCDEF = 353798