Puzzle for December 13, 2020 ( )
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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.
* D! is D-factorial. E! is E-factorial.
Many thanks to Tom H for submitting this interesting puzzle. Thank you, Tom!
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Hint #1
Add A, B, and C to both sides of eq.3: E – A – B + A + B + C = A – C + A + B + C which becomes eq.3a) E + C = 2×A + B Add D and E to both sides of eq.2: C – D + E + D + E = A + B – E + D + E which becomes C + 2×E = A + B + D which may be written as eq.2a) E + E + C = A + B + D
Hint #2
In eq.2a, replace E + C with 2×A + B (from eq.3a): E + 2×A + B = A + B + D Subtract A and B from both sides of the above equation: E + 2×A + B – A – B = A + B + D – A – B which simplifies to eq.2b) E + A = D
Hint #3
In eq.4, replace D with E + A (from eq.2b): B + E = A + C + E + A which becomes B + E = 2×A + C + E Subtract E from each side of the above equation: B + E – E = 2×A + C + E – E which becomes eq.4a) B = 2×A + C
Hint #4
In eq.3, substitute (2×A + C) for B (from eq.4a): E – A – (2×A + C) = A – C which is equivalent to E – A – 2×A – C = A – C which becomes E – 3×A – C = A – C Add 3×A and C to both sides of the above equation: E – 3×A – C + 3×A + C = A – C + 3×A + C which makes E = 4×A
Hint #5
Substitute 4×A for E in eq.2b: 4×A + A = D which makes 5×A = D
Hint #6
D-factorial is defined as: D! = 1 × 2 × 3 × ... × (D – 1) × D which could be written as D! = (D – 1)! × D In eq.5, replace D! with (D – 1)! × D: (D – 1)! × D ÷ (D × C) = B which is equivalent to (D – 1)! × D ÷ D ÷ C = B which becomes (D – 1)! ÷ C = B Multiply both sides of the above equation by C: (D – 1)! ÷ C × C = B × C which becomes eq.5a) (D – 1)! = B × C
Hint #7
Substitute (D – 1)! for B × C (from eq.5a) into eq.6: E! = (D – 1)! which means eq.6a) E = D – 1
Hint #8
Substitute D – 1 for E (from eq.6a) in eq.2b: D – 1 + A = D In the above equation, subtract D from each side, and add 1 to each side: D – 1 + A – D + 1 = D – D + 1 which makes A = 1 making D = 5×A = 5 × 1 = 5 E = 4×A = 4 × 1 = 4
Hint #9
In eq.4, substitute 4 for E, 1 for A, and 5 for D: B + 4 = 1 + C + 5 which becomes B + 4 = C + 6 Subtract 4 from each side of the above equation: B + 4 – 4 = C + 6 – 4 which makes eq.4a) B = C + 2
Hint #10
Substitute 4 for E, and C + 2 for B (from eq.4a) in eq.6: 4! = (C + 2) × C which becomes 24 = C² + 2×C Subtract 24 from each side of the above equation: 24 – 24 = C² + 2×C – 24 which becomes eq.6b) 0 = C² + 2×C – 24
Hint #11
eq.6b is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.6b yields: C = { (–1)×(2) ± sq.rt.[(2)² – (4 × (1) × (–24))] } ÷ (2 × (1)) which becomes C = {–2 ± sq.rt.(4 – (–96)} ÷ 2 which becomes C = {–2 ± sq.rt.(100)} ÷ 2 which becomes C = (–2 ± 10) ÷ 2 In the above equation, either: C = (–2 + 10) ÷ 2 = 8 ÷ 2 = 4 or C = (–2 – 10) ÷ 2 = –12 ÷ 2 = –6 Since C must be non-negative, then C ≠ –6 and therefore makes C = 4
Hint #12
Substitute 4 for C in eq.4a: B = 4 + 2 which makes B = 6
Solution
Substitute 1 for A, 6 for B, 4 for C and E, and 5 for D in eq.1: 1 + 6 + 4 + 5 + 4 + F = 20 which simplifies to 20 + F = 20 Subtract 20 from both sides of the above equation: 20 + F – 20 = 20 – 20 which makes F = 0 and makes ABCDEF = 164540