Puzzle for February 17, 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
Add A and D to both sides of eq.2: D - A + A + D = C - D + A + D which becomes eq.2a) 2×D = C + A Add D to both sides of eq.5: B + C - D + D = A + D + D which becomes eq.5a) B + C = A + 2×D
Hint #2
In eq.5a, replace 2×D with C + A (from eq.2a): B + C = A + C + A which becomes B + C = 2×A + C Subtract C from each side of the above equation: B + C - C = 2×A + C - C which makes B = 2×A
Hint #3
In eq.3, substitute 2×A for B: E - A = A - 2×A which becomes E - A = -A Add A to both sides of the equation above: E - A + A = -A + A which makes E = 0
Hint #4
eq.6 may be written as: C = (A + B + D + F) ÷ 4 Multiply both sides of the above equation by 4: 4 × C = 4 × (A + B + D + F) ÷ 4 which becomes eq.6a) 4×C = A + B + D + F
Hint #5
eq.1 may be re-written as: A + B + D + F + C + E = 25 In the above equation, substitute 4×C for A + B + D + F (from eq.6a), and 0 for E: 4×C + C + 0 = 25 which makes 5×C = 25 Divide both sides by 5: 5×C ÷ 5 = 25 ÷ 5 which makes C = 5
Hint #6
Substitute 2×A for B, and 5 for C in eq.5a: 2×A + 5 = A + 2×D Subtract A from each side of the equation above: 2×A + 5 - A = A + 2×D - A which becomes A + 5 = 2×D Divide both sides by 2: (A + 5) ÷ 2 = 2×D ÷ 2 which becomes eq.5b) ½×A + 2½ = D
Hint #7
Substitute ½×A + 2½ for D (from eq.5b), and 0 for E in eq.4: F - A = ½×A + 2½ - 0 which becomes F - A = ½×A + 2½ Add A to both sides of the above equation: F - A + A = ½×A + 2½ + A which becomes eq.4a) F = 1½×A + 2½
Hint #8
Substitute 5 for C, 2×A for B, ½×A + 2½ for D (from eq.5b), and 1½×A + 2½ for F (from eq.4a) in eq.6a: 4×5 = A + 2×A + ½×A + 2½ + 1½×A + 2½ which becomes 20 = 5×A + 5 Subtract 5 from each side of the above equation: 20 - 5 = 5×A + 5 - 5 which makes 15 = 5×A Divide both sides by 5: 15 ÷ 5 = 5×A ÷ 5 which means 3 = A
Solution
Since A = 3, then: B = 2×A = 2 × 3 = 6 D = ½×A + 2½ = ½×3 + 2½ = 1½ + 2½ = 4 (from eq.5b) F = 1½×A + 2½ = 1½×3 + 2½ = 4½ + 2½ = 7 (from eq.4a) and ABCDEF = 365407