Puzzle for December 1, 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.3, substitute (A + D) for E (from eq.2): D - A = C - (A + D) which becomes D - A = C - A - D Add A and D to both sides of the above equation: D - A + A + D = C - A - D + A + D which makes eq.3a) 2×D = C
Hint #2
Add the left and right sides of eq.3 to the left and right sides of eq.5, respectively: A + B + D - A = C + E - F + C - E which becomes eq.5a) B + D = 2×C - F
Hint #3
In eq.6, replace B + D with 2×C - F (from eq.5a): C + F = 2×C - F - F which becomes C + F = 2×C - 2×F In the above equation, subtract C from both sides, and add 2×F to both sides: C + F - C + 2×F = 2×C - 2×F - C + 2×F which makes 3×F = C
Hint #4
In eq.3a, replace C with 3×F: 2×D = 3×F Divide both sides of the above equation by 2: 2×D ÷ 2 = 3×F ÷ 2 which makes D = 1½×F
Hint #5
In eq.5a, substitute 1½×F for D, and (3×F) for C: B + 1½×F = 2×(3×F) - F which becomes B + 1½×F = 6×F - F which becomes B + 1½×F = 5×F Subtract 1½×F from each side of the equation above: B + 1½×F - 1½×F = 5×F - 1½×F which makes B = 3½×F
Hint #6
Substitute 1½×F for D, 3½×F for B, and 3×F for C in eq.4: 1½×F + F = A - 3½×F + 3×F which becomes 2½×F = A - ½×F Add ½×F to both sides of the equation above: 2½×F + ½×F = A - ½×F + ½×F which makes 3×F = A
Hint #7
Substitute 3×F for A, and 1½×F for D in eq.2: E = 3×F + 1½×F which becomes E = 4½×F
Solution
Substitute 3×F for A and C, 3½×F for B, 1½×F for D, and 4½×F for E in eq.1: 3×F + 3½×F + 3×F + 1½×F + 4½×F + F = 33 which simplifies to 16½×F = 33 Divide both sides of the above equation by 16½: 16½×F ÷ 16½ = 33 ÷ 16½ which means F = 2 making A = C = 3×F = 3 × 2 = 6 B = 3½×F = 3½ × 2 = 7 D = 1½×F = 1½ × 2 = 3 E = 4½×F = 4½ × 2 = 9 and ABCDEF = 676392