Puzzle for September 25, 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.
* CD and EF are 2-digit numbers (not C×D or E×F).
Scratchpad
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Hint #1
Subtract F and B from both sides of eq.3: D + F - F - B = A + B - F - B which becomes eq.3a) D - B = A - F
Hint #2
In eq.4, replace A - F with D - B (from eq.3a): F - D = D - B Add D and B to both sides of the above equation: F - D + D + B = D - B + D + B which becomes eq.4a) F + B = 2×D
Hint #3
Add B and C to both sides of eq.5: A - B - C + E + F + B + C = B + C + B + C which becomes A + E + F = 2×B + 2×C which may be written as eq.5a) E + A + F = 2×B + 2×C
Hint #4
eq.6 may be written as: B + 10×C + D - A = D + 10×E + F In the above equation, subtract D from both sides, and add A to both sides: B + 10×C + D - A - D + A = D + 10×E + F - D + A which becomes B + 10×C = 10×E + F + A which may be written as eq.6a) B + 10×C = 9×E + E + A + F
Hint #5
In eq.6a, replace E + A + F with 2×B + 2×C (from eq.5a): B + 10×C = 9×E + 2×B + 2×C Subtract B, 9×E, and 2×C from both sides of the above equation: B + 10×C - B - 9×E - 2×C = 9×E + 2×B + 2×C - B - 9×E - 2×C which simplifies to eq.6b) 8×C - 9×E = B
Hint #6
In eq.6a, substitute 8×C - 9×E for B (from eq.6b), and D + E for A + F (from eq.2): 8×C - 9×E + 10×C = 9×E + E + D + E which becomes 18×C - 9×E = 11×E + D Subtract 11×E from each side of the equation above: 18×C - 9×E - 11×E = 11×E + D - 11×E which becomes eq.6c) 18×C - 20×E = D
Hint #7
Substitute 8×C - 9×E for B (from eq.6b), and (18×C - 20×E) for D (from eq.6c) in eq.4a: F + 8×C - 9×E = 2×(18×C - 20×E) which becomes F + 8×C - 9×E = 36×C - 40×E In the above equation, subtract 8×C from both sides, and add 9×E to both sides: F + 8×C - 9×E - 8×C + 9×E = 36×C - 40×E - 8×C + 9×E which becomes eq.4b) F = 28×C - 31×E
Hint #8
Subtract the left and right sides of eq.3 from the left and right sides of eq.2, respectively: D + E - (D + F) = A + F - (A + B) which becomes D + E - D - F = A + F - A - B which becomes E - F = F - B Add F and B to both sides of the above equation: E - F + F + B = F - B + F + B which becomes eq.2a) B + E = 2×F
Hint #9
Substitute 8×C - 9×E for B (from eq.6b), and (28×C - 31×E) for F (from eq.4b) in eq.2a: 8×C - 9×E + E = 2×(28×C - 31×E) which becomes 8×C - 8×E = 56×C - 62×E which becomes 8×C - 8×E = 56×C - 62×E In the above equation, subtract 8×C from both sides, and add 62×E to both sides: 8×C - 8×E - 8×C + 62×E = 56×C - 62×E - 8×C + 62×E which makes 54×E = 48×C Divide both sides by 48: 54×E ÷ 48 = 48×C ÷ 48 which makes eq.2b) 1⅛×E = C
Hint #10
Substitute (1⅛×E) for C in eq.6b: 8×(1⅛×E) - 9×E = B which becomes 9×E - 9×E = B which means 0 = B
Hint #11
Substitute (1⅛×E) for C in eq.6c: 18×(1⅛×E) - 20×E = D which becomes 20¼×E - 20×E = D which makes ¼×E = D Multiply both sides of the above equation by 4: 4 × ¼×E = 4 × D which makes E = 4×D
Hint #12
Substitute 0 for B, and 4×D for E in eq.2a: 0 + 4×D = 2×F which makes 4×D = 2×F Divide both sides of the above equation by 2: 4×D ÷ 2 = 2×F ÷ 2 which makes 2×D = F
Hint #13
Substitute (4×D) for E in eq.2b: 1⅛×(4×D) = C which makes 4½×D = C
Hint #14
Substitute 2×D for F in eq.4: 2×D - D = A - 2×D which becomes D = A - 2×D Add 2×D to both sides of the above equation: D + 2×D = A - 2×D + 2×D which becomes 3×D = A
Solution
Substitute 3×D for A, 0 for B, 4½×D for C, 4×D for E, and 2×D for F in eq.1: 3×D + 0 + 4½×D + D + 4×D + 2×D = 29 which simplifies to 14½×D = 29 Divide both sides of the above equation by 14½: 14½×D ÷ 14½ = 29 ÷ 14½ which means D = 2 making A = 3×D = 3 × 2 = 6 C = 4½×D = 4½ × 2 = 9 E = 4×D = 4 × 2 = 8 F = 2×D = 2 × 2 = 4 and ABCDEF = 609284