Puzzle for September 27, 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 positive integer.
* A! is A-factorial.
Scratchpad
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Hint #1
Add C to both sides of eq.2: A + C + C = D + E + C which is the same as A + 2×C = C + D + E In the equation above, replace C + D with A + B (from eq.3): A + 2×C = A + B + E Subtract A from both sides: A + 2×C – A = A + B + E – A which becomes eq.2a) 2×C = B + E
Hint #2
In eq.1, replace B + E with 2×C (from eq.2a): eq.1a) D + F = 2×C
Hint #3
eq.5 may be written as: A × C = B + E + D + F In the above equation, substitute 2×C for B + E (from eq.2a) and for D + F (from eq.1a): A × C = 2×C + 2×C which becomes A × C = 4×C Divide both sides by C: A × C ÷ C = 4×C ÷ C which makes A = 4
Hint #4
eq.5 may also be written as: A × C = B + F + D + E In the equation above, substitute A + D for B + F (from eq.4), and A + C for D + E (from eq.2): A × C = A + D + A + C which becomes eq.5a) A × C = 2×A + D + C
Hint #5
Substitute 4 for A in eq.5a: 4 × C = 2×4 + D + C which becomes 4×C = 8 + D + C Subtract 8 and C from each side of the above equation: 4×C – 8 – C = 8 + D + C – 8 – C which becomes eq.5b) 3×C – 8 = D
Hint #6
Substitute 3×C – 8 for D (from eq.5b) in eq.1a: 3×C – 8 + F = 2×C In the above equation above, add 8 to both sides, and subtract 3×C from both sides: 3×C – 8 + F + 8 – 3×C = 2×C + 8 – 3×C which simplifies to eq.1b) F = 8 – C
Hint #7
Substitute 3×C – 8 for D (from eq.5b), and 4 for A in eq.3: C + 3×C – 8 = 4 + B which becomes 4×C – 8 = 4 + B Subtract 4 from each side of the above equation: 4×C – 8 – 4 = 4 + B – 4 which becomes eq.3a) 4×C – 12 = B
Hint #8
Substitute 4×C – 12 for B (from eq.3a) in eq.2a: 2×C = 4×C – 12 + E In the above equation, subtract 4×C from both sides, and add 12 to both sides: 2×C – 4×C + 12 = 4×C – 12 + E – 4×C + 12 which becomes –2×C + 12 = E which may be written as eq.2b) 12 – 2×C = E
Hint #9
Substitute 4 for A, 4×C – 12 for B (from eq.3a), and (12 – 2×C) for E (from eq.2b) in eq.6: 4! = (4 + 4×C – 12) × (12 – 2×C) which becomes 4 × 3 × 2 × 1 = (4×C – 8) × (12 – 2×C) which becomes 24 = 48×C – 8×C² – 96 + 16×C which becomes 24 = 64×C – 8×C² – 96 Subtract 24 from both sides of the above equation: 24 – 24 = 64×C – 8×C² – 96 – 24 which becomes 0 = 64×C – 8×C² – 120 Divide both sides by (–8): 0 ÷ (–8) = (64×C – 8×C² – 120) ÷ (–8) which becomes 0 = –8×C + C² + 15 which may be written as eq.6a) 0 = C² – 8×C + 15
Hint #10
eq.6a is a quadratic equation in standard form. Using the quadratic equation solution formula to solve for C in eq.6a yields: C = { (–1)×(–8) ± sq.rt.[(–8)² – (4 × (1) × (15))] } ÷ (2 × (1)) which becomes C = {8 ± sq.rt.(64 – 60)} ÷ 2 which becomes C = {8 ± sq.rt.(4)} ÷ 2 which becomes C = (8 ± 2) ÷ 2 In the above equation, either C = (8 + 2) ÷ 2 = 10 ÷ 2 = 5 or C = (8 – 2) ÷ 2 = 6 ÷ 2 = 3
Solution
Check C = 3 ... Substituting 3 for C in eq.3a would yield: B = 4×3 – 12 which would become B = 12 – 12 which would make B = 0 Since B must be a positive integer, then B ≠ 0 which means C ≠ 3 and therefore makes C = 5 making B = 4×C – 12 = 4×5 – 12 = 20 – 12 = 8 (from eq.3a) D = 3×C – 8 = 3×5 – 8 = 15 – 8 = 7 (from eq.5b) E = 12 – 2×C = 12 – 2×5 = 12 – 10 = 2 (from eq.2b) F = 8 – C = 8 – 5 = 3 (from eq.1b) and ABCDEF = 485723