Puzzle for July 16, 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.2, subtract D from both sides, and add F to both sides: B + D - D + F = E - F - D + F which becomes eq.2a) B + F = E - D In eq.5, subtract C from both sides, and add D and F to both sides: B + C - D - C + D + F = A - C + D - F - C + D + F which becomes eq.5a) B + F = A - 2×C + 2×D
Hint #2
In eq.5a, replace B + F with E - D (from eq.2a): E - D = A - 2×C + 2×D Add D and 2×C to both sides of the above equation: E - D + D + 2×C = A - 2×C + 2×D + D + 2×C which becomes E + 2×C = A + 3×D which may be written as eq.5b) E + 2×C = A + D + 2×D
Hint #3
Add D and E to both sides of eq.4: E - D + D + E = A - E + D + E which becomes eq.4a) 2×E = A + D
Hint #4
In eq.5b, replace A + D with 2×E (from eq.4a): E + 2×C = 2×E + 2×D Subtract E from each side of the equation above: E + 2×C - E = 2×E + 2×D - E which becomes 2×C = E + 2×D which may be written as eq.5c) 2×C = E + D + D
Hint #5
Add D and B to both sides of eq.3: C - D + D + B = E - B + D + B which becomes eq.3a) C + B = E + D
Hint #6
In eq.5c, substitute C + B for E + D (from eq.3a): 2×C = C + B + D Subtract C from each side of the equation above: 2×C - C = C + B + D - C which becomes eq.5d) C = B + D
Hint #7
Substitute B + D for C (from eq.5d) in eq.3a: B + D + B = E + D which becomes 2×B + D = E + D Subbtract D from both sides of the above equation: 2×B + D - D = E + D - D which makes eq.3b) 2×B = E
Hint #8
eq.6 may be written as: (E + F) ÷ 2 = (A + B + C + D) ÷ 4 Multiply both sides of the above equation by 4: 4 × (E + F) ÷ 2 = 4 × (A + B + C + D) ÷ 4 which becomes 2 × (E + F) = A + B + C + D which becomes eq.6a) 2×E + 2×F = A + B + C + D
Hint #9
Substitute 2×E + 2×F for A + B + C + D (from eq.6a) into eq.1: 2×E + 2×F + E + F = 18 which becomes 3×E + 3×F = 18 which may be written as 3×(E + F) = 18 Divide both sides of the above equation by 3: 3×(E + F) ÷ 3 = 18 ÷ 3 which becomes E + F = 6 Subtract F from each side: E + F - F = 6 - F which becomes eq.1a) E = 6 - F
Hint #10
Substitute 6 - F for E (from eq.1a) in eq.3b: 2×B = 6 - F Divide both sides of the above equation by 2: 2×B ÷ 2 = (6 - F) ÷ 2 which becomes eq.3c) B = 3 - ½×F
Hint #11
Substitute 3 - ½×F for B (from eq.3c) and 6 - F for E (from eq.1a) in eq.2: 3 - ½×F + D = 6 - F - F which becomes 3 - ½×F + D = 6 - 2×F In the above equation, subtract 3 from both sides, and add ½×F to both sides: 3 - ½×F + D - 3 + ½×F = 6 - 2×F - 3 + ½×F which becomes eq.2b) D = 3 - 1½×F
Hint #12
Substitute 3 - 1½×F for D (from eq.2b) and 3 - ½×F for B (from eq.3c) in eq.5d: C = 3 - ½×F + 3 - 1½×F which becomes eq.5e) C = 6 - 2×F
Hint #13
Substitute 6 - F for E (from eq.1a), and 3 - 1½×F for D (from eq.2b) in eq.4a: 2×(6 - F) = A + 3 - 1½×F which becomes 12 - 2×F = A + 3 - 1½×F In the equation above, subtract 3 from both sides, and add 1½×F to both sides: 12 - 2×F - 3 + 1½×F = A + 3 - 1½×F - 3 + 1½×F which becomes eq.4b) 9 - ½×F = A
Hint #14
In eq.1, substitute 9 - ½×F for A (from eq.4b), 3 - ½×F for B (from eq.3c), 6 - 2×F for C (from eq.5e), 3 - 1½×F for D (from eq.2b), and 6 - F for E (from eq.1a): 9 - ½×F + 3 - ½×F + 6 - 2×F + 3 - 1½×F + 6 - F + F = 18 which simplifies to 27 - 4½×F = 18 In the above equation, add 4½×F to both sides, and subtract 18 from both sides: 27 - 4½×F + 4½×F - 18 = 18 + 4½×F - 18 which makes 9 = 4½×F Divide both sides by 4½: 9 ÷ 4½ = 4½×F ÷ 4½ which means 2 = F
Solution
Since F = 2, then: A = 9 - ½×F = 9 - ½×2 = 9 - 1 = 8 (from eq.4b) B = 3 - ½×F = 3 - ½×2 = 3 - 1 = 2 (from eq.3c) C = 6 - 2×F = 6 - 2×2 = 6 - 4 = 2 (from eq.5e) D = 3 - 1½×F = 3 - 1½×2 = 3 - 3 = 0 (from eq.2b) E = 6 - F = 6 - 2 = 4 (from eq.1a) and ABCDEF = 822042