Puzzle for March 8, 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
In eq.5, replace B + C with F (from eq.3): E + F = A + F Subtract F from both sides of the above equation: E + F – F = A + F – F which makes E = A
Hint #2
Subtract C from each side of eq.5: E + F – C = A + B + C – C which becomes eq.5a) E + F – C = A + B Add C to each side of eq.2: C + F + C = A + B – C + C which becomes eq.2a) 2×C + F = A + B
Hint #3
In eq.5a, replace A + B with 2×C + F (from eq.2a): E + F – C = 2×C + F In the equation above, subtract F from both sides, and add C to both sides: E + F – C – F + C = 2×C + F – F + C which makes E = 3×C and also makes A = E = 3×C
Hint #4
In eq.6, substitute 3×C for both A and E: F – 3×C = 3×C × C Add 3×C to both sides of the above equation: F – 3×C + 3×C = (3×C × C) + 3×C which becomes eq.6a) F = 3×C² + 3×C
Hint #5
Substitute 3×C² + 3×C for F (from eq.6a) in eq.3: B + C = 3×C² + 3×C Subtract C from each side of the equation above: B + C – C = 3×C² + 3×C – C which becomes eq.3a) B = 3×C² + 2×C
Hint #6
Substitute 3×C for A, (3×C² + 2×C) for B (from eq.3a), and 3×C² + 3×C for F (from eq.6a) in eq.4: 3×C – (3×C² + 2×C) + C + 3×C² + 3×C = 3×C² + 2×C + D which becomes 7×C – 3×C² – 2×C + 3×C² = 3×C² + 2×C + D which becomes 5×C = 3×C² + 2×C + D Subtract 3×C² and 2×C from each side of the above equation: 5×C – 3×C² – 2×C = 3×C² + 2×C + D – 3×C² – 2×C which makes eq.4a) 3×C – 3×C² = D
Hint #7
Substitute 3×C for A and E, 3×C² + 2×C for B (from eq.3a), 3×C – 3×C² for D (from eq.4a), and 3×C² + 3×C for F (from eq.6a) in eq.1: 3×C + 3×C² + 2×C + C + 3×C – 3×C² + 3×C + 3×C² + 3×C = 18 which simplifies to 3×C² + 15×C = 18 Subtract 18 from each side of the above equation: 3×C² + 15×C – 18 = 18 – 18 which becomes eq.1a) 3×C² + 15×C – 18 = 0
Solution
eq.1a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.1a yields: C = {–15 ± sq.rt.(15² – 4 × 3 × (–18))} ÷ (2 × 3) which becomes C = {–15 ± sq.rt.(225 + 216)} ÷ 6 which becomes C = {–15 ± sq.rt.(441)} ÷ 6 which becomes C = (–15 ± 21) ÷ 6 In the above equation, either C = (–15 + 21) ÷ 6 = 6 ÷ 6 = 1 or C = (–15 – 21) ÷ 6 = –36 ÷ 6 = –6 Since C must be non-negative, then C ≠ –6 and therefore makes C = 1 making A = E = 3×C = 3 × 1 = 3 B = 3×C² + 2×C = 3×1² + 2×1 = 3 + 2 = 5 (from eq.3a) D = 3×C – 3×C² = 3×1 – 3×1² = 3 – 3 = 0 (from eq.4a) F = 3×C² + 3×C = 3×1² + 3×1 = 3 + 3 = 6 (from eq.6a) and ABCDEF = 351036