Puzzle for December 14, 2020 ( )
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
eq.4 may be written as: A + D = C + F + E In the above equation, replace C + F with D + E (from eq.6): A + D = D + E + E which becomes A + D = D + 2×E Subtract D from both sides: A + D – D = D + 2×E – D which makes A = 2×E
Hint #2
In eq.2, replace A with 2×E: C – E = 2×E + E which becomes C – E = 3×E Add E to both sides of the equation above: C – E + E = 3×E + E which makes C = 4×E
Hint #3
In eq.5, substitute 2×E for A, and 4×E for C: B + E = 2×E – B + 4×E which becomes B + E = 6×E – B In the above equation, subtract E from each side, and add B to each side: B + E – E + B = 6×E – B – E + B which becomes 2×B = 5×E Divide both sides by 2: 2×B ÷ 2 = 5×E ÷ 2 which becomes B = 2½×E
Hint #4
Substitute 2×E for A, and 2½×E for B in eq.3: D = 2×E + 2½×E which makes D = 4½×E
Hint #5
Substitute 4½×E for D, and 4×E for C in eq.6: 4½×E + E = 4×E + F which becomes 5½×E = 4×E + F Subtract 4×E from each side of the equation above: 5½×E – 4×E = 4×E + F – 4×E which makes 1½×E = F
Solution
Substitute 2×E for A, 2½×E for B, 4×E for C, 4½×E for D, and 1½×E for F in eq.1: 2×E + 2½×E + 4×E + 4½×E + E + 1½×E = 31 which simplifies to 15½×E = 31 Divide both sides of the equation above by 15½: 15½×E ÷ 15½ = 31 ÷ 15½ which means E = 2 making A = 2×E = 2 × 2 = 4 B = 2½×E = 2½ × 2 = 5 C = 4×E = 4 × 2 = 8 D = 4½×E = 4½ × 2 = 9 F = 1½×E = 1½ × 2 = 3 and ABCDEF = 458923