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 Bound the product (Posted on 2015-02-13)
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 No Rating

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 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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