Puzzle for October 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 positive integer.
* B! means B-factorial.
Scratchpad
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Hint #1
In eq.3, substitute (D + E) for F (from eq.1): E - (D + E) = B - C which becomes E - D - E = B - C which becomes -D = B - C Add D and C to both sides of the above equation: -D + D + C = B - C + D + C which becomes eq.3a) C = B + D
Hint #2
In eq.2, replace B + D with C (from eq.3a): A = C
Hint #3
In eq.4, substitute B + D for A (from eq.2), and D + E for F (from eq.1): B + E = B + D - B + D + E which becomes B + E = 2×D + E Subtract E from each side of the equation above: B + E - E = 2×D + E - E which makes B = 2×D
Hint #4
Substitute 2×D for B in eq.3a: C = 2×D + D which makes C = 3×D and also makes A = C = 3×D
Hint #5
In eq.5, substitute D + E for F (from eq.1), 2×D for B, and 3×D for A: D + E - 2×D - D = 3×D + 2×D - E which becomes E - 2×D = 5×D - E Add 2×D and E to both sides of the equation above: E - 2×D + 2×D + E = 5×D - E + 2×D + E which makes 2×E = 7×D Divide both sides by 2: 2×E ÷ 2 = 7×D ÷ 2 which makes E = 3½×D
Hint #6
Substitute 3½×D for E in eq.1: F = D + 3½×D which makes F = 4½×D
Hint #7
Substitute (2×D) for B, 3×D for C, 3½×D for E, and 4½×D for F in eq.6: (2×D)! = 3×D + D + 3½×D + 4½×D which becomes (2×D)! = 12×D The above equation may be written as: 1 × 2 × 3 × ... × (2×D – 2) × (2×D – 1) × (2×D) = 12×D which is the same as (2×D – 1)! × (2×D) = 12×D Divide both sides by (2×D): (2×D – 1)! × (2×D) ÷ (2×D) = 12×D ÷ (2×D) which becomes (2×D – 1)! = 6 which means eq.6a) (2×D – 1)! = 3!
Hint #8
eq.6a makes: (2×D – 1) = 3 Add 1 to both sides of the above equation: (2×D – 1) + 1 = 3 + 1 which makes 2×D = 4 Divide both sides by 2: 2×D ÷ 2 = 4 ÷ 2 which makes D = 2
Solution
Since D = 2, then A = C = 3×D = 3 × 2 = 6 B = 2×D = 2 × 2 = 4 E = 3½×D = 3½ × 2 = 7 F = 4½×D = 4½ × 2 = 9 and ABCDEF = 646279