Puzzle for February 20, 2022 ( )
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.
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Hint #1
Add A to both sides of eq.3: A + B + A = E – A + A which becomes eq.3a) 2×A + B = E In eq.5, replace E with 2×A + B (from eq.3a): A + D + F = B + 2×A + B – F which becomes A + D + F = 2×B + 2×A – F Subtract A and F from both sides of the above equation: A + D + F – A – F = 2×B + 2×A – F – A – F which becomes eq.5a) D = 2×B + A – 2×F
Hint #2
In eq.4, replace E with 2×A + B (from eq.3a), and C with B + D (from eq.2): 2×A + B + F = B + D + D which becomes 2×A + B + F = B + 2×D Subtract B from each side of the equation above: 2×A + B + F – B = B + 2×D – B which becomes 2×A + F = 2×D Divide both sides by 2: (2×A + F) ÷ 2 = 2×D ÷ 2 which becomes eq.4a) A + ½×F = D
Hint #3
In eq.4a, substitute 2×B + A – 2×F for D (from eq.5a): A + ½×F = 2×B + A – 2×F In the above equation, subtract A from both sides, and add 2×F to both sides: A + ½×F – A + 2×F = 2×B + A – 2×F – A + 2×F which becomes 2½×F = 2×B Divide both sides by 2: 2½×F ÷ 2 = 2×B ÷ 2 which makes 1¼×F = B
Hint #4
Substitute 1¼×F for B in eq.3a: eq.3b) 2×A + 1¼×F = E
Hint #5
Substitute 1¼×F for B, and A + ½×F for D (from eq.4a) in eq.2: C = 1¼×F + A + ½×F which becomes eq.2a) C = 1¾×F + A
Hint #6
Substitute 1¼×F for B, 1¾×F + A for C (from eq.2a), and 2×A + 1¼×F for E (from eq.3b) in eq.6: 1¼×F×F = A + 1¾×F + A + 2×A + 1¼×F which becomes 1¼×F² = 4×A + 3×F Subtract 3×F from each side of the equation above: 1¼×F² – 3×F = 4×A + 3×F – 3×F which becomes 1¼×F² – 3×F = 4×A Divide both sides by 4: (1¼×F² – 3×F) ÷ 4 = 4×A ÷ 4 which becomes eq.6a) 0.3125×F² – ¾×F = A
Hint #7
Substitute 0.3125×F² – ¾×F for A (from eq.6a) in eq.4a: 0.3125×F² – ¾×F + ½×F = D which becomes eq.4b) 0.3125×F² – ¼×F = D
Hint #8
Substitute 0.3125×F² – ¾×F for A (from eq.6a) in eq.2a: C = 1¾×F + 0.3125×F² – ¾×F which becomes eq.2b) C = F + 0.3125×F²
Hint #9
Substitute (0.3125×F² – ¾×F) for A (from eq.6a) in eq.3b: 2×(0.3125×F² – ¾×F) + 1¼×F = E which becomes 0.625×F² – 1½×F + 1¼×F = E which becomes eq.3c) 0.625×F² – ¼×F = E
Hint #10
In eq.1, substitute –– 0.3125×F² – ¾×F for A (from eq.6a), 1¼×F for B, F + 0.3125×F² for C (from eq.2b), 0.3125×F² – ¼×F for D (from eq.4b), and 0.625×F² – ¼×F for E (from eq.3c): 0.3125×F² – ¾×F + 1¼×F + F + 0.3125×F² + 0.3125×F² – ¼×F + 0.625×F² – ¼×F + F = 33 which simplifies to 1.5625×F² + 2×F = 33 Subtract 33 from both sides of the above equation: 1.5625×F² + 2×F – 33 = 33 – 33 which becomes eq.1a) 1.5625×F² + 2×F – 33 = 0
Hint #11
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)×(2) ± sq.rt.[(2)² – (4 × (1.5625) × (–33))] } ÷ (2 × (1.5625)) which becomes F = {–2 ± sq.rt.[4 – (–206.25)]} ÷ 3.125 which becomes F = {–2 ± sq.rt.[210.25]} ÷ 3.125 which becomes F = (–2 ± 14.5) ÷ 3.125 In the above equation, either F = (–2 + 14.5) ÷ 3.125 = 12.5 ÷ 3.125 = 4 or F = (–2 – 14.5) ÷ 3.125 = –16.5 ÷ 3.125 = –5.28 Since F must be a non-negative integer, then: F ≠ –5.28 and therefore makes: F = 4
Solution
Since F = 4, then: A = 0.3125×F² – ¾×F = 0.3125×4² – ¾×4 = 0.3125×16 – 3 = 5 – 3 = 2 (from eq.6a) B = 1¼×F = 1¼ × 4 = 5 C = 0.3125×F² + F = 0.3125×4² + 4 = 0.3125×16 + 4 = 5 + 4 = 9 (from eq.2b) D = 0.3125×F² – ¼×F = 0.3125×4² – ¼×4 = 0.3125×16 – 1 = 5 – 1 = 4 (from eq.4b) E = 0.625×F² – ¼×F = 0.625×4² – ¼×4 = 0.625×16 – 1 = 10 – 1 = 9 (from eq.3c) and ABCDEF = 259494