Puzzle for September 5, 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
In eq.3, replace E with A + B (from eq.1): A + B – D = A – B Subtract A from both sides of the above equation: A + B – D – A = A – B – A which becomes B – D = –B Add B and D to both sides: B – D + B + D = –B + B + D which makes 2×B = D
Hint #2
Add A to both sides of eq.4: C + E – A + A = A + D + A which becomes eq.4a) C + E = 2×A + D In eq.6, replace C + E with 2×A + D (from eq.4a): D = (A + 2×A + D) ÷ B which becomes D = (3×A + D) ÷ B Multiply both sides of the above equation by B: D × B = ((3×A + D) ÷ B) × B which becomes eq.6a) D × B = 3×A + D
Hint #3
In eq.6a, substitute 2×B for D: 2×B × B = 3×A + 2×B which may be written as 2×B² = 3×A + 2×B Subtract 2×B from each side of the equation above: 2×B² – 2×B = 3×A + 2×B – 2×B which becomes 2×B² – 2×B = 3×A Divide both sides by 3: (2×B² – 2×B) ÷ 3 = 3×A ÷ 3 which becomes eq.6b) ⅔×B² – ⅔×B = A
Hint #4
Substitute ⅔×B² – ⅔×B for A (from eq.6b) in eq.1: ⅔×B² – ⅔×B + B = E which becomes eq.1a) ⅔×B² + ⅓×B = E
Hint #5
Substitute ⅔×B² + ⅓×B for E (from eq.1a), (⅔×B² – ⅔×B) for A (from eq.6b), and 2×B for D in eq.4a: C + ⅔×B² + ⅓×B = 2×(⅔×B² – ⅔×B) + 2×B which becomes C + ⅔×B² + ⅓×B = 1⅓×B² – 1⅓×B + 2×B which becomes C + ⅔×B² + ⅓×B = 1⅓×B² + ⅔×B Subtract ⅔×B² and ⅓×B from both sides of the equation above: C + ⅔×B² + ⅓×B – ⅔×B² – ⅓×B = 1⅓×B² + ⅔×B – ⅔×B² – ⅓×B which makes C = ⅔×B² + ⅓×B and also makes eq.1b) C = E = ⅔×B² + ⅓×B (from eq.1a)
Hint #6
Substitute ⅔×B² + ⅓×B for C (from eq.1b), and ⅔×B² – ⅔×B for A (from eq.6b) in eq.2: B + ⅔×B² + ⅓×B = ⅔×B² – ⅔×B + F which becomes ⅔×B² + 1⅓×B = ⅔×B² – ⅔×B + F In the above equation, subtract ⅔×B² from both sides, and add ⅔×B to both sides: ⅔×B² + 1⅓×B – ⅔×B² + ⅔×B = ⅔×B² – ⅔×B + F – ⅔×B² + ⅔×B which simplifies to 2×B = F
Hint #7
Substitute 2×B for D and F in eq.5: A = (2×B + 2×B) ÷ B which becomes A = 4×B ÷ B which becomes A = 4
Hint #8
Substitute 4 for A in eq.6b: ⅔×B² – ⅔×B = 4 Multiply both sides of the above equation by 1½: (⅔×B² – ⅔×B) × 1½ = 4 × 1½ which becomes B² – B = 6 Subtract 6 from both sides: B² – B – 6 = 6 – 6 which becomes eq.6c) B² – B – 6 = 0
Solution
eq.6c is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for B in eq.6c yields: B = { (–1)×(–1) ± sq.rt.[(–1)² – (4 × (1) × (–6))] } ÷ (2 × (1)) which becomes B = {1 ± sq.rt.(1 – (–24))} ÷ 2 which becomes B = {1 ± sq.rt.(25)} ÷ 2 which becomes B = (1 ± 5) ÷ 2 In the above equation, either B = (1 + 5) ÷ 2 = 6 ÷ 2 = 3 or B = (1 – 5) ÷ 2 = –4 ÷ 2 = –2 Since B must be non-negative, then B ≠ –2 and therefore makes B = 3 and makes C = E = ⅔×B² + ⅓×B = ⅔×3² + ⅓×3 = ⅔×9 + 1 = 6 + 1 = 7 (from eq.1b) D = F = 2×B = 2×3 = 6 and ABCDEF = 437676