Puzzle for January 11, 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
In eq.5, add A and F to both sides, and subtract C from both sides: F - A + A + F - C = C - F + A + F - C which becomes eq.5a) 2×F - C = A Add C to both sides of eq.2: B - C + C = C - F + C which becomes eq.2a) B = 2×C - F
Hint #2
In eq.6, substitute (2×F - C) for A (from eq.5a), and 2×C - F for B (from eq.2a): (2×F - C) + F = 2×C - F + C - (2×F - C) which becomes 3×F - C = 3×C - F - 2×F + C which becomes 3×F - C = 4×C - 3×F Add C and 3×F to both sides of the above equation: 3×F - C + C + 3×F = 4×C - 3×F + C + 3×F which makes eq.6a) 6×F = 5×C
Hint #3
Multiply both sides of eq.5a by 3: 3×(2×F - C) = 3×A which becomes 6×F - 3×C = 3×A In the above equation, replace 6×F with 5×C (from eq.6a): 5×C - 3×C = 3×A which becomes 2×C = 3×A Divide both sides by 2: 2×C ÷ 2 = 3×A ÷ 2 which makes C = 1½×A
Hint #4
In eq.6a, substitute (1½×A) for C: 6×F = 5×(1½×A) which becomes 6×F = 7½×A Divide both sides of the above equation by 6: 6×F ÷ 6 = 7½×A ÷ 6 which makes F = 1¼×A
Hint #5
In eq.2a, substitute (1½×A) for C, and 1¼×A for F: B = 2×(1½×A) - 1¼×A which becomes B = 3×A - 1¼×A which makes B = 1¾×A
Hint #6
Substitute 1½×A for C in eq.3: E - A = A - 1½×A which becomes E - A = -½×A Add A to both sides of the above equation: E - A + A = -½×A + A which makes E = ½×A
Hint #7
Substitute 1½×A for C, and ½×A for E in eq.4: 1½×A - ½×A = A - D which becomes A = A - D Subtract A from each side of the equation above: A - A = A - D - A which makes 0 = -D which means 0 = D
Solution
Substitute 1¾×A for B, 1½×A for C, 0 for D, ½×A for E, and 1¼×A for F in eq.1: A + 1¾×A + 1½×A + 0 + ½×A + 1¼×A = 24 which simplifies to 6×A = 24 Divide both sides of the above equation by 6: 6×A ÷ 6 = 24 ÷ 6 which means A = 4 making B = 1¾×A = 1¾ × 4 = 7 C = 1½×A = 1½ × 4 = 6 E = ½×A = ½ × 4 = 2 F = 1¼×A = 1¼ × 4 = 5 and ABCDEF = 476025