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Some Factorials Equal Score Exponent (Posted on 2024-04-04) Difficulty: 3 of 5
Find all possible values of a real number r such that:
r! = r20

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

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Some Thoughts integral value and real -- a sampling | Comment 1 of 3
Searching factorials to 100! finds no integral solutions, except for 1^20 = 1!., but we can search for the generalized factorial over the reals.

First trying for integral solution, with standard factorials:

for r=20:20:100
  disp([r factorial(r) r^20])
end

narrowed this down to between 20 and 40.

Then

for r=sym(20):40
  f=factorial(r);
  p=r^20;
  disp(r)
  disp(f)
  disp(p)
  disp(' ')
end


showed that the passing point was between 27 and 28.  Neither integral value had an equality.

The generalized factorial is the Gamma function of n+1, so:

low=27; high=28; dx=.01;
found=false; prev=0;
while ~found
for i=low-dx:dx:high
  diff=genFact(i)-i^20;
  if diff==0
    found=true;
    break
  end
  if i>low-dx/2
    if prev*diff<0
      low=i-dx;
      high=i;
      dx=dx/100;
      break
    end
  end
  prev=diff;
end
end

function v=genFact(n)
  v=gamma(n+1);
end


>> rFactorialEqScoreExponent
Operation terminated by user during rFactorialEqScoreExponent 

>> i
i =
          27.5247695102146           this is r
>> gamma(i+1)
ans =
      6.22979589706583e+28          and the equal values
>> i^20
ans =
      6.22979589706581e+28           of the functions
 
 

I haven't explored negative r, but for example, (-2)^20 = 1048576 and (-3)^20 = 3486784401. In between these x values, the Gamma function of x+1 has two arms extending without limit upward from about two and a third. There must be two crossing points in that range. I did just one of these, r=-2.00000095366484 where each function evaluates to about 1048586.

  Posted by Charlie on 2024-04-04 09:38:59
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