Puzzle for March 18, 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 the left and right sides of eq.4 to the left and right sides of eq.5, respectively: C + E - F + C - D - E = A - E + F + D + E - F which becomes 2×C - F - D = A + D Add F and D to both sides of the above equation: 2×C - F - D + F + D = A + D + F + D which becomes eq.5a) 2×C = A + 2×D + F
Hint #2
Add D and C to both sides of eq.6: C - D + E + D + C = A - C + D + F + D + C which becomes eq.6a) 2×C + E = A + 2×D + F
Hint #3
In eq.6a, replace A + 2×D + F with 2×C (from eq.5a): 2×C + E = 2×C Subtract 2×C from each side of the equation above: 2×C + E - 2×C = 2×C - 2×C which makes E = 0
Hint #4
In eq.5, substitute 0 for E: C + 0 - F = A - 0 + F which becomes C - F = A + F Add F to both sides of the above equation: C - F + F = A + F + F which becomes eq.5b) C = A + 2×F
Hint #5
In eq.4, substitute 0 for E: C - D - 0 = D + 0 - F which becomes C - D = D - F Add D and F to both sides of the equation above: C - D + D + F = D - F + D + F which becomes eq.4a) C + F = 2×D
Hint #6
Substitute 0 for E, and 2×D for C + F (from eq.4a) into eq.3: B + D + 0 = A + 2×D which becomes B + D = A + 2×D Subtract D from each side of the above equation: B + D - D = A + 2×D - D which becomes eq.3a) B = A + D
Hint #7
Substitute A + D for B (from eq.3a) in eq.2: D - A = A + D - C In the above equation, subtract D from both sides, and add A and C to both sides: D - A - D + A + C = A + D - C - D + A + C which simplifies to eq.2a) C = 2×A
Hint #8
Substitute 2×A for C in eq.5b: 2×A = A + 2×F Subtract A from each side of the above equation: 2×A - A = A + 2×F - A which makes A = 2×F
Hint #9
Substitute (2×F) for A in eq.2a: C = 2×(2×F) which makes C = 4×F
Hint #10
Substitute 4×F for C in eq.4a: 4×F + F = 2×D which makes 5×F = 2×D Divide both sides of the above equation by 2: 5×F ÷ 2 = 2×D ÷ 2 which makes 2½×F = D
Hint #11
Substitute 2×F for A, and 2½×F for D in eq.3a: B = 2×F + 2½×F which makes B = 4½×F
Solution
Substitute 2×F for A, 4½×F for B, 4×F for C, 2½×F for D, and 0 for E in eq.1: 2×F + 4½×F + 4×F + 2½×F + 0 + F = 28 which simplifies to 14×F = 28 Divide both sides of the above equation by 14: 14×F ÷ 14 = 28 ÷ 14 which means F = 2 making A = 2×F = 2 × 2 = 4 B = 4½×F = 4½ × 2 = 9 C = 4×F = 4 × 2 = 8 D = 2½×F = 2½ × 2 = 5 and ABCDEF = 498502