Puzzle for October 14, 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.6 may be written as: C = (A + D + E) ÷ 3 Multiply both sides of the above equation by 3: 3 × C = 3 × (A + D + E) ÷ 3 which becomes 3×C = A + D + E which may be written as eq.6a) 3×C = A + E + D
Hint #2
In eq.6a, replace A + E with C + D (from eq.3): 3×C = C + D + D which becomes 3×C = C + 2×D Subtract C from each side of the equation above: 3×C - C = C + 2×D - C which makes 2×C = 2×D Divide both sides by 2: 2×C ÷ 2 = 2×D ÷ 2 which makes C = D
Hint #3
eq.5 may be written as: A = (B + C + F) ÷ 3 Multiply both sides of the above equation by 3: 3 × A = 3 × (B + C + F) ÷ 3 which becomes eq.5a) 3×A = B + C + F
Hint #4
eq.1 may be written as: A + E + D + B + C + F = 36 In the above equation, replace A + E + D with 3×C (from eq.6a), and B + C + F with 3×A (from eq.5a): 3×C + 3×A = 36 Divide both sides by 3: (3×C + 3×A) ÷ 3 = 36 ÷ 3 which becomes C + A = 12 Subtract C from both sides: C + A - C = 12 - C which becomes eq.1a) A = 12 - C
Hint #5
In eq.2, substitute D for C: E - B = B - D Add B and D to both sides of the above equation: E - B + B + D = B - D + B + D which becomes eq.2a) E + D = 2×B
Hint #6
Substitute 12 - C for A (from eq.1a), and 2×B for E + D (from eq.2a) into eq.6a: 3×C = 12 - C + 2×B In the equation above, subtract 12 from both sides, and add C to both sides: 3×C - 12 + C = 12 - C + 2×B - 12 + C which becomes eq.6b) 4×C - 12 = 2×B Divide both sides of eq.6b by 2: (4×C - 12) ÷ 2 = 2×B ÷ 2 which becomes eq.6c) 2×C - 6 = B
Hint #7
Substitute C for D, and 4×C - 12 for 2×B (from eq.6b) in eq.2a: E + C = 4×C - 12 Subtract C from each side of the above equation: E + C - C = 4×C - 12 - C which becomes eq.2b) E = 3×C - 12
Hint #8
Substitute (12 - C) for A (from eq.1a), and 2×C - 6 for B (from eq.6c) in eq.5a: 3×(12 - C) = 2×C - 6 + C + F which becomes 36 - 3×C = 3×C - 6 + F In the above equation, subtract 3×C from both sides, and add 6 to both sides: 36 - 3×C - 3×C + 6 = 3×C - 6 + F - 3×C + 6 which becomes eq.5b) 42 - 6×C = F
Hint #9
In eq.4, substitute -- 2×C - 6 for B (from eq.6c), (12 - C) for A (from eq.1a), (42 - 6×C) for F (from eq.5b), C for D, and (3×C - 12) for E (from eq.2b): 2×C - 6 - (12 - C) + (42 - 6×C) = (12 - C) + C - (3×C - 12) - (42 - 6×C) which becomes 2×C - 6 - 12 + C + 42 - 6×C = 12 - C + C - 3×C + 12 - 42 + 6×C which becomes 24 - 3×C = -18 + 3×C Add 3×C and 18 to both sides of the equation above: 24 - 3×C + 3×C + 18 = -18 + 3×C + 3×C + 18 which becomes 42 = 6×C Divide both sides by 6: 42 ÷ 6 = 6×C ÷ 6 which makes 7 = C
Solution
Since C = 7, then: A = 12 - C = 12 - 7 = 5 (from eq.1a) B = 2×C - 6 = 2×7 - 6 = 14 - 6 = 8 (from eq.6c) D = C = 7 E = 3×C - 12 = 3×7 - 12 = 21 - 12 = 9 (from eq.2b) F = 42 - 6×C = 42 - 6×7 = 42 - 42 = 0 (from eq.5b) and ABCDEF = 587790