Puzzle for September 2, 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.
* AB, BC, and DE are 2-digit numbers (not A×B, B×C, or D×E).
Scratchpad
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Hint #1
In eq.4, add C and E to both sides, and subtract D from both sides: D - C + C + E - D = C - E + C + E - D which becomes eq.4a) E = 2×C - D
Hint #2
In eq.4a, replace E with B + D (from eq.2): B + D = 2×C - D Subtract D from both sides of the above equation: B + D - D = 2×C - D - D which becomes eq.4b) B = 2×C - 2×D
Hint #3
In eq.3, replace B with 2×C - 2×D (from eq.4b): 2×C - 2×D + F = A + D Subtract D from each side of the equation above: 2×C - 2×D + F - D = A + D - D which becomes eq.3a) 2×C - 3×D + F = A
Hint #4
In eq.5, substitute 2×C - 3×D + F for A (from eq.3a), 2×C - 2×D for B (from eq.4b), and 2×C - D for E (from eq.4a): 2×C - 3×D + F + F = 2×C - 2×D + C + 2×C - D which becomes 2×C - 3×D + 2×F = 5×C - 3×D In the above equation, subtract 2×C from both sides, and add 3×D to both sides: 2×C - 3×D + 2×F - 2×C + 3×D = 5×C - 3×D - 2×C + 3×D which simplifies to 2×F = 3×C Divide both sides by 2: 2×F ÷ 2 = 3×C ÷ 2 which makes eq.5a) F = 1½×C
Hint #5
Substitute 1½×C for F in eq.3a: 2×C - 3×D + 1½×C = A which becomes eq.3b) 3½×C - 3×D = A
Hint #6
eq.6 may be written as: 10×A + B = 10×B + C + 10×D + E Subtract B from each side of the equation above: 10×A + B - B = 10×B + C + 10×D + E - B which becomes eq.6a) 10×A = 9×B + C + 10×D + E
Hint #7
Substitute (3½×C - 3×D) for A (from eq.3b), (2×C - 2×D) for B (from eq.4b), and 2×C - D for E (from eq.4a) in eq.6a: 10×(3½×C - 3×D) = 9×(2×C - 2×D) + C + 10×D + 2×C - D which becomes 35×C - 30×D = 18×C - 18×D + C + 10×D + 2×C - D which becomes 35×C - 30×D = 21×C - 9×D In the above equation, add 30×D to both sides, and subtract 21×C from both sides: 35×C - 30×D + 30×D - 21×C = 21×C - 9×D + 30×D - 21×C which simplifies to 14×C = 21×D Divide both sides by 14: 14×C ÷ 14 = 21×D ÷ 14 which makes C = 1½×D
Hint #8
Substitute (1½×D) for C in eq.3b: 3½×(1½×D) - 3×D = A which becomes 5¼×D - 3×D = A which makes 2¼×D = A
Hint #9
Substitute (1½×D) for C in eq.4b: B = 2×(1½×D) - 2×D which becomes B = 3×D - 2×D which makes B = D
Hint #10
Substitute (1½×D) for C in eq.4a: E = 2×(1½×D) - D which becomes E = 3×D - D which makes E = 2×D
Hint #11
Substitute (1½×D) for C in eq.5a: F = 1½×(1½×D) which makes F = 2¼×D
Solution
Substitute 2¼×D for A and F, D for B, 1½×D for C, and 2×D for E in eq.1: 2¼×D + D + 1½×D + D + 2×D + 2¼×D = 40 which simplifies to 10×D = 40 Divide both sides of the above equation by 10: 10×D ÷ 10 = 40 ÷ 10 which means D = 4 making A = F = 2¼×D = 2¼ × 4 = 9 B = D = 4 C = 1½×D = 1½ × 4 = 6 E = 2×D = 2 × 4 = 8 and ABCDEF = 946489