Puzzle for September 11, 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.
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Hint #1
eq.2 may be written as: (A + F) ÷ 2 = C Multiply both sides of the above equation by 2: 2 × (A + F) ÷ 2 = 2 × C which becomes A + F = 2×C Subtract F from both sides: A + F - F = 2×C - F which becomes eq.2a) A = 2×C - F
Hint #2
eq.6 may be written as: (A + C + D + E) ÷ 4 = B - D Multiply both sides of the above equation by 4: 4 × (A + C + D + E) ÷ 4 = 4 × (B - D) which becomes A + C + D + E = 4×B - 4×D Add 4×D to both sides: A + C + D + E + 4×D = 4×B - 4×D + 4×D which becomes eq.6a) A + C + 5×D + E = 4×B
Hint #3
In eq.6a, replace A with 2×C - F (from eq.2a): 2×C - F + C + 5×D + E = 4×B which becomes 3×C - F + 5×D + E = 4×B Add F to both sides of the above equation: 3×C - F + 5×D + E + F = 4×B + F which becomes eq.6b) 3×C + 5×D + E = 4×B + F
Hint #4
eq.3 may be written as: (B + E) ÷ 2 = D + E Multiply both sides of the above equation by 2: 2 × (B + E) ÷ 2 = 2 × (D + E) which becomes B + E = 2×D + 2×E Subtract E from both sides: B + E - E = 2×D + 2×E - E which becomes eq.3a) B = 2×D + E
Hint #5
In eq.6b, substitute (2×D + E) for B (from eq.3a): 3×C + 5×D + E = 4×(2×D + E) + F which becomes 3×C + 5×D + E = 8×D + 4×E + F Subtract 5×D and E from each side of the equation above: 3×C + 5×D + E - 5×D - E = 8×D + 4×E + F - 5×D - E which becomes eq.6c) 3×C = 3×D + 3×E + F
Hint #6
eq.4 may be written as: (D + E + F) ÷ 3 = A - E Multiply both sides of the above equation by 3: 3 × (D + E + F) ÷ 3 = 3 × (A - E) which becomes D + E + F = 3×A - 3×E Add 3×E to both sides: D + E + F + 3×E = 3×A - 3×E + 3×E which becomes eq.4a) D + 4×E + F = 3×A
Hint #7
In eq.4a, substitute (2×C - F) for A (from eq.2a): D + 4×E + F = 3×(2×C - F) which becomes D + 4×E + F = 6×C - 3×F which may be written as eq.4b) D + 4×E + F = 2×(3×C) - 3×F
Hint #8
Substitute 3×D + 3×E + F for 3×C (from eq.6c) into eq.4b: D + 4×E + F = 2×(3×D + 3×E + F) - 3×F which becomes D + 4×E + F = 6×D + 6×E + 2×F - 3×F which becomes D + 4×E + F = 6×D + 6×E - F In the equation above, subtract D and 4×E from both sides, and add F to both sides: D + 4×E + F - D - 4×E + F = 6×D + 6×E - F - D - 4×E + F which simplifies to eq.4c) 2×F = 5×D + 2×E
Hint #9
eq.5 may be written as: (B + C + E) ÷ 3 = F - D Multiply both sides of the above equation by 3: 3 × (B + C + E) ÷ 3 = 3 × (F - D) which becomes eq.5a) B + C + E = 3×F - 3×D
Hint #10
Substitute 2×D + E for B (from eq.3a) in eq.5a: 2×D + E + C + E = 3×F - 3×D which becomes 2×D + 2×E + C = 3×F - 3×D Subtract 2×D and 2×E from each side of the above equation: 2×D + 2×E + C - 2×D - 2×E = 3×F - 3×D - 2×D - 2×E which becomes eq.5b) C = 3×F - 5×D - 2×E
Hint #11
Substitute 3×F - 5×D - 2×E for C (from eq.5b) in eq.6c: 3×(3×F - 5×D - 2×E) = 3×D + 3×E + F which becomes 9×F - 15×D - 6×E = 3×D + 3×E + F In the above equation, add 15×D and 6×E to both sides, and subtract F from both sides: 9×F - 15×D - 6×E + 15×D + 6×E - F = 3×D + 3×E + F + 15×D + 6×E - F which simplifies to 8×F = 18×D + 9×E which may be written as eq.6d) 4×(2×F) = 18×D + 9×E
Hint #12
Substitute 5×D + 2×E for 2×F (from eq.4c) in eq.6d: 4×(5×D + 2×E) = 18×D + 9×E which becomes 20×D + 8×E = 18×D + 9×E Subtract 8×E and 18×D from each side of the equation above: 20×D + 8×E - 8×E - 18×D = 18×D + 9×E - 8×E - 18×D which simplifies to 2×D = E
Hint #13
Substitute 2×D for E in eq.3a: B = 2×D + 2×D which makes B = 4×D
Hint #14
Substitute (2×D) for E in eq.4c: 2×F = 5×D + 2×(2×D) which becomes 2×F = 5×D + 4×D which makes 2×F = 9×D Divide both sides of the above equation by 2: 2×F ÷ 2 = 9×D ÷ 2 which makes F = 4½×D
Hint #15
Substitute (4½×D) for F, and (2×D) for E in eq.5b: C = 3×(4½×D) - 5×D - 2×(2×D) which becomes C = 13½×D - 5×D - 4×D which makes C = 4½×D
Hint #16
Substitute (4½×D) for C and F in eq.2a: A = 2×(4½×D) - (4½×D) which becomes A = 9×D - 4½×D which makes A = 4½×D
Solution
Substitute 4½×D for A and C and F, 4×D for B, and 2×D for E in eq.1: 4½×D + 4×D + 4½×D + D + 2×D + 4½×D = 41 which simplifies to 20½×D = 41 Divide both sides of the above equation by 20½: 20½×D ÷ 20½ = 41 ÷ 20½ which means D = 2 making A = C = F = 4½×D = 4½ × 2 = 9 B = 4×D = 4 × 2 = 8 E = 2×D = 2 × 2 = 4 and ABCDEF = 989249