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A Pair of Factorial Quartic Equations (Posted on 2022-07-02) Difficulty: 3 of 5
Solve for n (a positive integer) in each equation:
(1) n! = 8n^4 + 15n^3 - 4n^2 + 15n + 8
(2) n! = 9n^4 + 4n^3 + n^2 + 1344

Both analytic and computer solutions welcome.

  Submitted by Larry    
Rating: 4.0000 (2 votes)
Solution: (Hide)
The solution to both equations is 8.

If you divide both sides of either equation by n you can eliminate all but a few choices for n
(1) (n-1)! = 8n^3 + 15n^2 - 4n^1 + 15 + 8/n
and
(2) (n-1)! = 9n^3 + 4n^2 + n + 1344/n

And knowing that the final term must be an integer limits the possibilties to divisors of 8 for equation 1 and 1355 for equation 2.
For the second equation, n! must be greater than 1344, so n > 6 for that one.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
No Subjecthoadao34932022-07-23 13:27:10
Solutionvery computerized solutionCharlie2022-07-02 10:32:34
SolutionSemi-Analytic Puzzle Solution: Part 2K Sengupta2022-07-02 07:23:22
SolutionSemi-Analytic Puzzle Solution: Part 1K Sengupta2022-07-02 07:00:11
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