Puzzle for July 13, 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.5, replace C with A + B (from eq.2): A + B - E = A + D In the above equation, subtract A from both sides, and add E to both sides: A + B - E - A + E = A + D - A + E which becomes eq.5a) B = D + E
Hint #2
In eq.4, substitute (D + E) for B (from eq.5a): (D + E) - D = C - (D + E) which becomes E = C - D - E Add D and E to both sides of the above equation: E + D + E = C - D - E + D + E which becomes eq.4a) 2×E + D = C
Hint #3
eq.6 may be written as: D = (A + B + C + E) ÷ 4 Multiply both sides of the above equation by 4: 4 × D = 4 × (A + B + C + E) ÷ 4 which becomes eq.6a) 4×D = A + B + C + E
Hint #4
In eq.6a, replace A + B with C (from eq.2): 4×D = C + C + E which becomes eq.6b) 4×D = 2×C + E
Hint #5
In eq.6b, substitute (2×E + D) for C (from eq.4a): 4×D = 2×(2×E + D) + E which becomes 4×D = 4×E + 2×D + E which becomes 4×D = 5×E + 2×D Subtract 2×D from each side of the equation above: 4×D - 2×D = 5×E + 2×D - 2×D which makes 2×D = 5×E Divide both sides by 2: 2×D ÷ 2 = 5×E ÷ 2 which makes D = 2½×E
Hint #6
Substitute 2½×E for D in eq.4a: 2×E + 2½×E = C which makes 4½×E = C
Hint #7
Substitute 2½×E for D in eq.5a: B = 2½×E + E which makes B = 3½×E
Hint #8
Substitute 2½×E for D in eq.3: 2½×E = E + F Subtract E from each side of the above equation: 2½×E - E = E + F - E which makes 1½×E = F
Hint #9
Substitute 4½×E for C, and 3½×E for B in eq.2: 4½×E = A + 3½×E Subtract 3½×E from both sides of the above equation: 4½×E - 3½×E = A + 3½×E - 3½×E which makes E = A
Solution
Substitute E for A, 3½×E for B, 4½×E for C, 2½×E for D, and 1½×E for F in eq.1: E + 3½×E + 4½×E + 2½×E + E + 1½×E = 28 which simplifies to 14×E = 28 Divide both sides of the above equation by 14: 14×E ÷ 14 = 28 ÷ 14 which means E = 2 making A = E = 2 B = 3½×E = 3½ × 2 = 7 C = 4½×E = 4½ × 2 = 9 D = 2½×E = 2½ × 2 = 5 F = 1½×E = 1½ × 2 = 3 and ABCDEF = 279523