Puzzle for December 30, 2018 ( )
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.
* AB and CD are 2-digit numbers (not A×B or C×D).
Scratchpad
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Hint #1
Add D + F to both sides of eq.4: B - D + D + F = E - F + D + F which becomes B + F = E + D which may be written as eq.4a) B + F = D + E In eq.5, replace D + E with B + F: B + B + F = A + F Subtract F from both sides of the equation above: B + B + F - F = A + F - F which makes 2×B = A
Hint #2
Add A + F to both sides of eq.3: D - A + A + F = A + C - F + A + F which becomes eq.3a) D + F = 2×A + C In eq.2, substitute 2×B for A: F - B + C = 2×B + B + E which becomes F - B + C = 3×B + E Subtract both 3×B and F from each side of the equation above: F - B + C - 3×B - F = 3×B + E - 3×B - F which becomes eq.2a) C - 4×B = E - F
Hint #3
In eq.4, substitute C - 4×B for E - F (from eq.2a): B - D = C - 4×B Add 4×B + D to both sides of the equation above: B - D + 4×B + D = C - 4×B + 4×B + D which becomes eq.4b) 5×B = C + D Subtract C from both sides of eq.4b: 5×B - C = C + D - C which becomes eq.4c) 5×B - C = D
Hint #4
Add F to each side of eq.4b: 5×B + F = C + D + F Substitute 2×A + C for D + F (from eq.3a): 5×B + F = C + 2×A + C which becomes 5×B + F = 2×A + 2×C Substitute (2×B) for A: 5×B + F = 2×(2×B) + 2×C which becomes 5×B + F = 4×B + 2×C Subtract 4×B from each side: 5×B + F - 4×B = 4×B + 2×C - 4×B which reduces to eq.2b) B + F = 2×C Substitute D + E for B + F (from eq.4a): D + E = 2×C Subtract D from each side: D + E - D = 2×C - D eq.2c) E = 2×C - D
Hint #5
eq.6 may be written as: 10×A + B = 10×C + D - E - F Add F to each side of the equation above: 10×A + B + F = 10×C + D - E - F + F which becomes 10×A + B + F = 10×C + D - E Substitute (2×B) for A, 2×C for B + F (from eq.2b), and (2×C - D) for E (from eq.2c): 10×(2×B) + 2×C = 10×C + D - (2×C - D) which becomes 20×B + 2×C = 10×C + D - 2×C + D which becomes 20×B + 2×C = 8×C + 2×D Subtract 2×C from each side: 20×B + 2×C - 2×C = 8×C + 2×D - 2×C which becomes eq.6a) 20×B = 6×C + 2×D
Hint #6
Substitute (5×B - C) for D (from eq.4c) in eq.6a: 20×B = 6×C + 2×(5×B - C) which is equivalent to 20×B = 6×C + 10×B - 2×C which becomes 20×B = 4×C + 10×B Subtract 10×B from each side: 20×B - 10×B = 4×C + 10×B - 10×B which simplifies to 10×B = 4×C Divide each side by 4: 10×B ÷ 4 = 4×C ÷ 4 which makes 2½×B = C
Hint #7
In eq.4c, replace C with 2½×B: 5×B - 2½×B = D which makes 2½×B = D
Hint #8
In eq.2c, replace both C and D with 2½×B: E = 2×(2½×B) - 2½×B which becomes E = 5×B - 2½×B which means E = 2½×B
Hint #9
Substitute 2×B for A, and 2½×B for D and E in eq.5: B + 2½×B + 2½×B = 2×B + F which becomes 6×B = 2×B + F Subtract 2×B from both sides of the above equation: 6×B - 2×B = 2×B + F - 2×B which makes 4×B = F
Solution
Substitute 2×B for A, 2½×B for C and D and E, and 4×B for F in eq.1: 2×B + B + 2½×B + 2½×B + 2½×B + 4×B = 29 which simplifies to 14½×B = 29 Divide both sides of the above equation by 14½: 14½×B ÷ 14½ = 29 ÷ 14½ which makes B = 2 making A = 2×B = 2 × 2 = 4 C = D = E = 2½×B = 2½ × 2 = 5 F = 4×B = 4 × 2 = 8 and ABCDEF = 425558