Puzzle for September 24, 2023 ( )
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
eq.4 may be written as: D + E - A = A + C - B In the equation above, replace A + C with D + F (from eq.3): D + E - A = D + F - B Subtract D from both sides, and add A and B to both sides: D + E - A - D + A + B = D + F - B - D + A + B which becomes E + B = F + A which may be written as eq.4a) B + E = A + F
Hint #2
In eq.2, replace A + F with B + E (from eq.4a): B + D = B + E Subtract B from each side of the equation above: B + D - B = B + E - B which makes D = E
Hint #3
eq.6 may be written as: C × D = A + F + E In the above equation, substitute B + D for A + F (from eq.2), and D for E: C × D = B + D + D which becomes C × D = B + 2×D Subtract 2×D from both sides: (C × D) - 2×D = B + 2×D - 2×D which becomes eq.6a) (C - 2) × D = B
Hint #4
In eq.5, substitute D for E: C + D - B = A - C + D In the above equation, subtract D from both sides, and add C to both sides: C + D - B - D + C = A - C + D - D + C which becomes eq.5a) 2×C - B = A
Hint #5
Subtract the left and right sides of eq.2 from the left and right sides of eq.3, respectively: D + F - (B + D) = A + C - (A + F) which becomes D + F - B - D = A + C - A - F which becomes F - B = C - F Add B and F to both sides of the above equation: F - B + B + F = C - F + B + F which becomes 2×F = C + B Divide both sides by 2: 2×F ÷ 2 = (C + B) ÷ 2 which becomes eq.3a) F = ½×C + ½×B
Hint #6
Substitute D for E, 2×C - B for A (from eq.5a), and ½×C + ½×B for F (from eq.3a) in eq.4a: B + D = 2×C - B + ½×C + ½×B which becomes B + D = 2½×C - ½×B Subtract B from both sides of the equation above: B + D - B = 2½×C - ½×B - B which makes D = 2½×C - 1½×B and also makes eq.4b) D = E = 2½×C - 1½×B
Hint #7
Substitute (2½×C - 1½×B) for D (from eq.4b) in eq.6a: (C - 2) × (2½×C - 1½×B) = B which becomes 2½×C² - 5×C - 1½×B×C + 3×B = B In the above equation, add 1½×B×C to both sides, and subtract 3×B from both sides: 2½×C² - 5×C - 1½×B×C + 3×B + 1½×B×C - 3×B = B + 1½×B×C - 3×B which becomes 2½×C² - 5×C = 1½×B×C - 2×B which may be written as 2½×C² - 5×C = (1½×C - 2)×B Divide both sides by (1½×C - 2): (2½×C² - 5×C) ÷ (1½×C - 2) = (1½×C - 2)×B ÷ (1½×C - 2) which becomes eq.6b) (2½×C² - 5×C) ÷ (1½×C - 2) = B
Hint #8
Substitute 2×C - B for A (from eq.5a), 2½×C - 1½×B for D and E (from eq.4b), and ½×C + ½×B for F (from eq.3a) in eq.1: 2×C - B + B + C + 2½×C - 1½×B + 2½×C - 1½×B + ½×C + ½×B = 17 which becomes eq.1a) 8½×C - 2½×B = 17
Hint #9
Substitute ((2½×C² - 5×C) ÷ (1½×C - 2)) for B (from eq.6b) in eq.1a: 8½×C - 2½×((2½×C² - 5×C) ÷ (1½×C - 2)) = 17 which becomes 8½×C - ((6¼×C² - 12½×C) ÷ (1½×C - 2)) = 17 In the above equation, add ((6¼×C² - 12½×C) ÷ (1½×C - 2)) to both sides, and subtract 17 from both sides: 8½×C - ((6¼×C² - 12½×C) ÷ (1½×C - 2)) + ((6¼×C² - 12½×C) ÷ (1½×C - 2)) - 17 = 17 + ((6¼×C² - 12½×C) ÷ (1½×C - 2)) - 17 which becomes eq.1b) 8½×C - 17 = (6¼×C² - 12½×C) ÷ (1½×C - 2)
Hint #10
Multiply both sides of eq.1b by (1½×C - 2): (1½×C - 2) × (8½×C - 17) = (1½×C - 2) × (6¼×C² - 12½×C) ÷ (1½×C - 2) which becomes 12¾×C² - 25½×C - 17×C + 34 = 6¼×C² - 12½×C which becomes 12¾×C² - 42½×C + 34 = 6¼×C² - 12½×C In the above equation, subtract 6¼×C² from both sides, and add 12½×C to both sides: 12¾×C² - 42½×C + 34 - 6¼×C² + 12½×C = 6¼×C² - 12½×C - 6¼×C² + 12½×C which becomes eq.1c) 6½×C² - 30×C + 34 = 0
Hint #11
eq.1c is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.1c yields: C = { (-1)×(-30) ± sq.rt.[(-30)² - (4 × 6½ × 34)] } ÷ (2 × 6½) which becomes C = {30 ± sq.rt.(900 - 884)} ÷ 13 which becomes C = {30 ± sq.rt.(16)} ÷ 13 which becomes eq.1d) C = (30 ± 4) ÷ 13
Hint #12
In eq.1d, either: C = (30 + 4) ÷ 13 = 34 ÷ 13 = 2.6153846153846 or: C = (30 - 4) ÷ 13 = 26 ÷ 13 = 2 Since C must be an integer, then C ≠ 2.6153846153846 and therefore makes C = 2
Hint #13
Substitute 2 for C in eq.6b: (2½×2² - 5×2) ÷ (1½×2 - 2) = B which becomes (2½×4 - 10) ÷ (3 - 2) = B which becomes (10 - 10) ÷ 1 = B which becomes 0 ÷ 1 = B which makes 0 = B
Hint #14
Substitute 2 for C, and 0 for B in eq.5a: 2×2 - 0 = A which makes 4 = A
Hint #15
Substitute 2 for C, and 0 for B in eq.4b: D = E = 2½×2 - 1½×0 which becomes D = E = 5 - 0 which makes D = E = 5
Solution
Substitute 2 for C, and 0 for B in eq.3a: F = ½×2 + ½×0 which becomes F = 1 + 0 which makes F = 1 and makes ABCDEF = 402551