Puzzle for December 10, 2022 ( )
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.
* DE is a 2-digit number (not D×E).
Scratchpad
Help Area
Hint #1
eq.4 may be written as: eq.4a) A + F - C = B + D - E In eq.3, subtract C and E from both sides, and add D to both sides: C + D - E - C - E + D = F - D + E - C - E + D which becomes eq.3a) 2×D - 2×E = F - C
Hint #2
In eq.4a, replace F - C with 2×D - 2×E (from eq.3a): A + 2×D - 2×E = B + D - E In the above equation, add 2×E to both sides, and subtract D from both sides: A + 2×D - 2×E + 2×E - D = B + D - E + 2×E - D which becomes eq.4b) A + D = B + E
Hint #3
In eq.2, replace A + D with B + E (from eq.4b): B + E = B + E - E which becomes B + E = B Subtract B from each side of the equation above: B + E - B = B - B which makes E = 0
Hint #4
In eq.3a, substitute 0 for E: 2×D - 2×0 = F - C which becomes eq.3b) 2×D = F - C Add C to both sides of eq.3b: 2×D + C = F - C + C which becomes eq.3c) 2×D + C = F
Hint #5
eq.6 may be written as: A ÷ D = (C + D + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × (A ÷ D) = 3 × (C + D + F) ÷ 3 which becomes eq.6a) 3×(A ÷ D) = C + D + F
Hint #6
eq.5 may be written as: (10×D + E) ÷ A = F - C Substitute 0 for E, and 2×D for F - C (from eq.3b) in the above equation: (10×D + 0) ÷ A = 2×D which becomes 10×D ÷ A = 2×D Multiply both sides by A, and, since D ≠ 0 (from eq.6), divide both sides by (2×D): ((10×D) ÷ A) × A ÷ (2×D) = 2×D × A ÷ (2×D) which simplifies to 5×D = A
Hint #7
Substitute 5×D for A, and 0 for E in eq.4b: 5×D + D = B + 0 which makes 6×D = B
Hint #8
Substitute (5×D) for A, and 2×D + C for F (from eq.3c) in eq.6a: 3×((5×D) ÷ D) = C + D + 2×D + C which becomes 15 = 2×C + 3×D Subtract 3×D from each side of the above equation: 15 - 3×D = 2×C + 3×D - 3×D which becomes 15 - 3×D = 2×C Divide both sides by 2: (15 - 3×D) ÷ 2 = 2×C ÷ 2 which makes eq.6b) 7½ - 1½×D = C
Hint #9
Substitute 7½ - 1½×D for C (from eq.6b) into eq.3c: 2×D + 7½ - 1½×D = F which makes eq.3d) ½×D + 7½ = F
Hint #10
Substitute 5×D for A, 6×D for B, 7½ - 1½×D for C (from eq.6b), 0 for E, and ½×D + 7½ for F (from eq.3d) in eq.1: 5×D + 6×D + 7½ - 1½×D + D + 0 + ½×D + 7½ = 26 which becomes 11×D + 15 = 26 Subtract 15 from each side of the above equation: 11×D + 15 - 15 = 26 - 15 which makes 11×D = 11 Divide both sides of the above equation by 11: 11×D ÷ 11 = 11 ÷ 11 which means D = 1
Solution
Since D = 1, then: A = 5×D = 5 × 1 = 5 B = 6×D = 6 × 1 = 6 C = 7½ - 1½×D = 7½ - (1½ × 1) = 7½ - 1½ = 6 (from eq.6b) F = ½×D + 7½ = (½ × 1) + 7½ = ½ + 7½ = 8 (from eq.3d) and ABCDEF = 566108