Puzzle for October 22, 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
Add C and F to both sides of eq.3: A - C + C + F = B + C - F + C + F which becomes eq.3a) A + F = B + 2×C
Hint #2
In eq.2, replace A + F with B + 2×C (from eq.3a): B + 2×C = B + C Subtract B and C from each side of the equation above: B + 2×C - B - C = B + C - B - C which makes C = 0
Hint #3
In eq.2, substitute 0 for C: B + 0 = A + F which becomes eq.2a) B = A + F
Hint #4
In eq.5, substitute 0 for C: A = (0 + E) ÷ F which becomes A = E ÷ F Multiply both sides of the above equation by F: A × F = (E ÷ F) × F which becomes eq.5a) A × F = E
Hint #5
In eq.6, replace B with A + F (from eq.2a): D = (A + F + E - F) ÷ A which becomes D = (A + E) ÷ A Multiply both sides of the above equation by A: A × D = A × (A + E) ÷ A which becomes A × D = A + E Subtract A from each side: A × D - A = A + E - A which becomes A × D - A = E which may be written as eq.6a) A × (D - 1) = E
Hint #6
Substitute A × (D - 1) for E (from eq.6a) in eq.5a: A × F = A × (D - 1) Since A ≠ 0 (from eq.6), divide both sides of the above equation by A: A × F ÷ A = A × (D - 1) ÷ A which makes eq.5b) F = D - 1
Hint #7
eq.4 may be written as: D = (C + E + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × D = 3 × (C + E + F) ÷ 3 which becomes eq.4a) 3×D = C + E + F
Hint #8
Substitute 0 for C, and D - 1 for F (from eq.5b) in eq.4a: 3×D = 0 + E + D - 1 which becomes 3×D = E + D - 1 In the above equation, subtract D from both sides, and add 1 to both sides: 3×D - D + 1 = E + D - 1 - D + 1 which becomes eq.4b) 2×D + 1 = E
Hint #9
Substitute (D - 1) for F (from eq.5b), and 2×D + 1 for E (from eq.4b) in eq.5a: A × (D - 1) = 2×D + 1 Divide both sides of the above equation by (D - 1): (assumes D ≠ 1) A × (D - 1) ÷ (D - 1) = (2×D + 1) ÷ (D - 1) which becomes eq.5c) A = (2×D + 1) ÷ (D - 1)
Hint #10
Confirm: D ≠ 1 ... If D = 1, then substituting 1 for D in eq.5b would yield: F = 1 - 1 which would make F = 0 Since F ≠ 0 (from eq.5), then: D ≠ 1
Hint #11
Substitute (2×D + 1) ÷ (D - 1) for A (from eq.5c), and D - 1 for F (from eq.5b) in eq.2a: eq.2b) B = (2×D + 1) ÷ (D - 1) + D - 1
Hint #12
Substitute ((2×D + 1) ÷ (D - 1)) for A (from eq.5c), ((2×D + 1) ÷ (D - 1)) + D - 1 for B (from eq.2b), 0 for C, 2×D + 1 for E (for eq.4b), and D - 1 for F (from eq.5b) in eq.1: ((2×D + 1) ÷ (D - 1)) + ((2×D + 1) ÷ (D - 1)) + D - 1 + D + 0 + 2×D + 1 + D - 1 = 25 which becomes (4×D + 2) ÷ (D - 1) + 5×D - 1 = 25 In the above equation, subtract 5×D from both sides, and add 1 to both sides: (4×D + 2) ÷ (D - 1) + 5×D - 1 - 5×D + 1 = 25 - 5×D + 1 which becomes eq.1a) (4×D + 2) ÷ (D - 1) = 26 - 5×D
Hint #13
Multiply both sides of eq.1a by (D - 1): (D - 1) × (4×D + 2) ÷ (D - 1) = (D - 1) × (26 - 5×D) which becomes 4×D + 2 = 26×D - 26 - 5×D² + 5×D which becomes 4×D + 2 = 31×D - 26 - 5×D² Subtract 4×D and 2 from each side of the equation above: 4×D + 2 - 4×D - 2 = 31×D - 26 - 5×D² - 4×D - 2 which becomes 0 = 27×D - 28 - 5×D² which may be written as eq.1b) 0 = -5×D² + 27×D - 28
Hint #14
eq.1b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for D in eq.1b yields: D = { (-1)×(27) ± sq.rt.[(27)² - (4 × (-5) × (-28))] } ÷ (2 × (-5)) which becomes D = {-27 ± sq.rt.(729 - 560)} ÷ (-10) which becomes D = {-27 ± sq.rt.(169)} ÷ (-10) which becomes eq.1c) D = (-27 ± 13) ÷ (-10)
Hint #15
In eq.1c, either: D = (-27 + 13) ÷ (-10) = -14 ÷ (-10) = 1.4 or: D = (-27 - 13) ÷ (-10) = -40 ÷ (-10) = 4 Since D must be an integer, then D ≠ 1.4 and therefore makes D = 4
Solution
Since D = 4, then: A = (2×D + 1) ÷ (D - 1) = (2×4 + 1) ÷ (4 - 1) = (8 + 1) ÷ 3 = 9 ÷ 3 = 3 (from eq.5c) B = (2×D + 1) ÷ (D - 1) + D - 1 = (2×4 + 1) ÷ (4 - 1) + 4 - 1 = (8 + 1) ÷ 3 + 3 = 9 ÷ 3 + 3 = 3 + 3 = 6 (from eq.2b) E = 2×D + 1 = 2×4 + 1 = 8 + 1 = 9 (from eq.4b) F = D - 1 = 4 - 1 = 3 (from eq.5b) and ABDCEF = 360493