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Bound the product (Posted on 2015-02-13) Difficulty: 2 of 5
Let
pr=(1/2)*(3/4)*(5/6)... *(99/100)

Without explicit computation prove that
pr is less than 1/10, but more than 1/15.

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts Some facts that may aid in finding a proof | Comment 1 of 4
The number is (100!/(50!*2^50)) / (50!*2^50), which reduces to 100! / ((50!)^2 * 2^100).

For what it's worth in guiding a solution to the problem, the computed value is:

 12611418068195524166851562157/158456325028528675187087900672 
 
The prime factors of the numerator (which is reduced to lowest terms) are:
 
 3  3  3  3  11  13  17  19  29  31  53  59  61  67  71  73  79  83  89  97 

The prime factors of the denominator are all 2's:
 
 2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2  2 

The value is approximately

 0.0795892373871787614 or  1/12.564512901854901014 
 
    5   open "bondprod.txt" for output as #2
   10   V=1
   20   for I=1 to 99 step 2
   30    V=V*I//(I+1)
   40   next
   50   print #2, V
   60   N=num(V)
   70   while N>1
   80      Pd=prmdiv(N):N=N//Pd
   90      print #2, Pd;
  100   wend
  110   print #2,:print #2,
  160   N=den(V)
  170   while N>1
  180      Pd=prmdiv(N):N=N//Pd
  190      print #2, Pd;
  200   wend
  210   print #2,:print #2,
  220   print #2, V/1,1/V

  Posted by Charlie on 2015-02-13 10:49:00
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