Puzzle for April 18, 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.2, replace E with A + C (from eq.3): B + C = A + A + C – C which becomes eq.2a) B + C = 2×A
Hint #2
In eq.4, replace B + C with 2×A (from eq.2a): eq.4a) D + E + F = 2×A
Hint #3
In eq.1, replace B + C with 2×A (from eq.2a), and D + E + F with 2×A (from eq.4a): A + 2×A + 2×A = 25 which becomes 5×A = 25 Divide both sides of the above equation by 5: 5×A ÷ 5 = 25 ÷ 5 which makes A = 5
Hint #4
In eq.3, substitute 5 for A: eq.3a) E = 5 + C
Hint #5
In eq.5, substitute 5 + C for E (from eq.3a): C + F = 5 + C – C which becomes C + F = 5 Subtract C from both sides of the equation above: C + F – C = 5 – C which makes eq.5a) F = 5 – C
Hint #6
Substitute 5 + C for E (from eq.3a), and 5 – C for F (from eq.5a), and 5 for A in eq.4a: D + 5 + C + 5 – C = 2×5 which becomes D + 10 = 10 Subtract 10 from both sides of the above equation: D + 10 – 10 = 10 – 10 which means D = 0
Hint #7
Substitute 5 for A in eq.2a: B + C = 2×5 which becomes B + C = 10 Subtract C from each side of the equation above: B + C – C = 10 – C which becomes eq.2b) B = 10 – C
Hint #8
Substitute (10 – C) for B (from eq.2b), (5 – C) for F (from eq.5a), (5 + C) for E (from eq.3a), and 5 for A in eq.6: ((10 – C) × (5 – C)) – (5 + C) = 5 × (5 + C) which becomes 50 – 5×C – 10×C + C² – 5 – C = 25 + 5×C which becomes 45 – 16×C + C² = 25 + 5×C Subtract 25 and 5×C from both sides of the equation above: 45 – 16×C + C² – 25 – 5×C = 25 + 5×C – 25 – 5×C which becomes 20 – 21×C + C² = 0 which may be written as eq.6a) C² – 21×C + 20 = 0
Solution
eq.6a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.6a yields: C = { (–1)×(–21) ± sq.rt.[(–21)² – (4 × 1 × 20)] } ÷ (2 × 1) which becomes C = {21 ± sq.rt.[441 – 80]} ÷ 2 which becomes C = {21 ± sq.rt.(361)} ÷ 2 which becomes C = {21 ± 19} ÷ 2 In the above equation, either C = (21 + 19) ÷ 2 = 40 ÷ 2 = 20 or C = (21 – 19) ÷ 2 = 2 ÷ 2 = 1 Because C must be a one-digit integer, then C ≠ 20 and therefore makes C = 1 making B = 10 – C = 10 – 1 = 9 (from eq.2b) E = 5 + C = 5 + 1 = 6 (from eq.3a) F = 5 – C = 5 – 1 = 4 (from eq.5a) and ABCDEF = 591064