Puzzle for September 4, 2021 ( )
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.
* C! is C-factorial.
Scratchpad
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Hint #1
Add E to both sides of eq.5: B – E + E = C + F – B + E which becomes B = C + F – B + E which may be written as eq.5a) B = C + E + F – B In eq.5a, replace C + E with A + B – C (from eq.3): B = A + B – C + F – B which becomes B = A – C + F which may be written as eq.3a) B = A + F – C
Hint #2
Add D and F to both sides of eq.4: A – D + D + F = C + D – F + D + F which becomes eq.4a) A + F = C + 2×D In eq.3a, replace A + F with C + 2×D (from eq.4a): B = C + 2×D – C which makes B = 2×D
Hint #3
In eq.2, substitute 2×D for B: 2×D – D = A – 2×D + F which becomes D = A – 2×D + F Add 2×D to both sides of the equation above: D + 2×D = A – 2×D + F + 2×D which becomes eq.2a) 3×D = A + F
Hint #4
In eq.4a, substitute 3×D for A + F (from eq.2a): 3×D = C + 2×D Subtract 2×D from both sides of the above equation: 3×D – 2×D = C + 2×D – 2×D which makes D = C
Hint #5
In eq.5a, substitute 2×D for B, and D for C: 2×D = D + E + F – 2×D which becomes 2×D = E + F – D Add D to each side of the above equation: 2×D + D = E + F – D + D which becomes eq.5b) 3×D = E + F
Hint #6
Substitute E + F for 3×D (from eq.5b) into eq.2a: E + F = A + F Subtract F from each side of the equation above: E + F – F = A + F – F which makes E = A
Hint #7
eq.6 may be written as: C! = A + F + B + D In the above equation, substitute D for C, 3×D for A + F (from eq.2a), and 2×D for B: D! = 3×D + 2×D + D which becomes D! = 6×D which may be written as eq.6a) D × (D – 1)! = 6×D
Hint #8
Divide both sides of eq.6a by D: D × (D – 1)! ÷ D = 6×D ÷ D which makes (D – 1)! = 6 which means D – 1 = 3 Add 1 to both sides of the above equation: D – 1 + 1 = 3 + 1 which makes D = 4 and makes B = 2×D = 2 × 4 = 8 C = D = 4
Hint #9
Substitute 4 for D in eq.2a: 3×4 = A + F which becomes 12 = A + F Subtract F from each side of the equation above: 12 – F = A + F – F which makes 12 – F = A and also makes eq.2b) E = A = 12 – F
Solution
Substitute 12 – F for A and E (from eq.2b), 8 for B, and 4 for C and D in eq.1: 12 – F + 8 + 4 + 4 + 12 – F + F = 37 which simplifies to 40 – F = 37 In the equation above, add F to both sides, and subtract 37 from both sides: 40 – F + F – 37 = 37 + F – 37 which means 3 = F making A = E = 12 – F = 12 – 3 = 9 (from eq.2b) and ABCDEF = 984493