Puzzle for March 22, 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.
Scratchpad
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Hint #1
In eq.4, replace F with B + C (from eq.3): C + B + C = A + B – C which becomes 2×C + B = A + B – C In the above equation, subtract B from both sides, and add C to both sides: 2×C + B – B + C = A + B – C – B + C which makes 3×C = A
Hint #2
eq.6 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.6a) 3×A = B + C + F
Hint #3
Substitute (3×C) for A, and F for B + C (from eq.3) in eq.6a: 3×(3×C) = F + F which makes 9×C = 2×F Divide both sides of the above equation by 2: 9×C ÷ 2 = 2×F ÷ 2 which makes 4½×C = F
Hint #4
Substitute 4½×C for F in eq.3: 4½×C = B + C Subtract C from both sides of the above equation: 4½×C – C = B + C – C which makes 3½×C = B
Hint #5
Substitute 3½×C for B, 3×C for A, and 4½×C for F in eq.5: 3½×C – C + E = 3×C + 4½×C – 3½×C which becomes 2½×C + E = 4×C Subtract 2½×C from each side of the above equation: 2½×C + E – 2½×C = 4×C – 2½×C which makes E = 1½×C
Hint #6
Substitute 3×C for A, and 1½×C for E in eq.2: 3×C = D + 1½×C Subtract 1½×C from each side of the equation above: 3×C – 1½×C = D + 1½×C – 1½×C which makes 1½×C = D
Solution
Substitute 3×C for A, 3½×C for B, 1½×C for D and E, and 4½×C for F in eq.1: 3×C + 3½×C + C + 1½×C + 1½×C + 4½×C = 30 which simplifies to 15×C = 30 Divide both sides of the above equation by 15: 15×C ÷ 15 = 30 ÷ 15 which means C = 2 making A = 3×C = 3 × 2 = 6 B = 3½×C = 3½ × 2 = 7 D = E = 1½×C = 1½ × 2 = 3 F = 4½×C = 4½ × 2 = 9 and ABCDEF = 672339