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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.

See The Solution Submitted by Larry    
Rating: 4.0000 (2 votes)

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Solution very computerized solution | Comment 3 of 4 |
Asking Wolfram Alpha to

solve n! = 8n^4 + 15n^3 - 4n^2 + 15n + 8

and

solve n! = 9n^4 + 4n^3 + n^2 + 1344

in each instance gives the solution in integers to be 8.

Verification in a Matlab session, results in 40320 for each of the LHS and RHS's:

>> n=8
n =
     8
>> 9n^4 + 4n^3 + n^2 + 1344
 9n^4 + 4n^3 + n^2 + 1344
  ?
Invalid expression. Check for missing multiplication operator,
missing or unbalanced delimiters, or other syntax error. To
construct matrices, use brackets instead of parentheses.
 
Did you mean:
>> 9*n^4 + 4*n^3 + n^2 + 1344
ans =
       40320
       
(It was easier to allow Matlab to supply the *'s than to insert them myself.)       
       
>> factorial(n)
ans =
       40320
>> 8n^4 + 15n^3 - 4n^2 + 15n + 8
 8n^4 + 15n^3 - 4n^2 + 15n + 8
  ?
Invalid expression. Check for missing multiplication operator,
missing or unbalanced delimiters, or other syntax error. To
construct matrices, use brackets instead of parentheses.
 
Did you mean:
>> 8*n^4 + 15*n^3 - 4*n^2 + 15*n + 8
ans =
       40320


  Posted by Charlie on 2022-07-02 10:32:34
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