Puzzle for November 3, 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.6, replace A + B with E + F (from eq.3): E + F - E = D + E + F - A which becomes F = D + E + F - A In the above equation, subtract F from both sides, and add A to both sides: F - F + A = D + E + F - A - F + A which becomes eq.6a) A = D + E
Hint #2
In eq.4, replace A with D + E (from eq.6a): C + D = D + E + E which becomes C + D = D + 2×E Subtract D from both sides of the above equation: C + D - D = D + 2×E - D which makes C = 2×E
Hint #3
In eq.2, replace C with 2×E, and D + E with A (from eq.6a): eq.2a) B + 2×E = A
Hint #4
In eq.5, substitute B + 2×E for A (from eq.2a), and 2×E for C: B + F = B + 2×E + 2×E which becomes B + F = B + 4×E Subtract B from each side of the above equation: B + F - B = B + 4×E which becomes F = 4×E
Hint #5
In eq.3, substitute 4×E for F, and B + 2×E for A (from eq.2a): E + 4×E = B + 2×E + B which becomes 5×E = 2×B + 2×E Subtract 2×E from both sides of the equation above: 5×E - 2×E = 2×B + 2×E - 2×E which makes 3×E = 2×B Divide both sides by 2: 3×E ÷ 2 = 2×B ÷ 2 which makes 1½×E = B
Hint #6
Substitute 1½×E for B, and 2×E for C in eq.2: 1½×E + 2×E = D + E which becomes 3½×E = D + E Subtract E from both sides of the above equation: 3½×E - E = D + E - E which makes 2½×E = D
Hint #7
Substitute 2×E for C, and 2½×E for D in eq.4: 2×E + 2½×E = A + E which becomes 4½×E = A + E Subtract E from each side of the above equation: 4½×E - E = A + E - E which makes 3½×E = A
Solution
Substitute 3½×E for A, 1½×E for B, 2×E for C, 2½×E for D, and 4×E for F in eq.1: 3½×E + 1½×E + 2×E + 2½×E + E + 4×E = 29 which simplifies to 14½×E = 29 Divide both sides of the above equation by 14½: 14½×E ÷ 14½ = 29 ÷ 14½ which means E = 2 making A = 3½×E = 3½ × 2 = 7 B = 1½×E = 1½ × 2 = 3 C = 2×E = 2 × 2 = 4 D = 2½×E = 2½ × 2 = 5 F = 4×E = 4 × 2 = 8 and ABCDEF = 734528