Hi Mohsen,

The highest number than can be represented using 31 bits is 2147483647 (2^31-1). In order to generate a random, 31-bit integer number, i.e. an integer in the range 0 to 2147483647, you would use RandomInteger[{0, 2^31 - 1}].

I hope this helps,

Cheers,

Edwin

Hi Mr.Bennink,

OK, yes, thank you.

But mathematica6.0 can represent numbers using more than 31 bits. Maybe this note that you said, is right & this is one bog for program of mathematica6.0. Of course, it is not problem. Can I use RandomInteger[]+2*RandomInteger[{2^29, 2^30}] instead of  RandomInteger[{2^30,2^31}] , that rule RandomInteger[] represent 0 or 1.

Do you mean that RandomInteger[{0, 2^31 - 1}] gives you integer number that are occasionally higher than 2^31-1? That would indeed be a bug.

The suggestion that you give would return numbers in the range 1073741824 (2*2^29) to 2147483649 (1+2*2^30).

A more intuitive way to generate a random 31-bit number, albeit not the fastest, is to first generate a list with 31 binary digits, and then convert it to an integer number:

FromDigits[RandomInteger[{0, 1}, 31], 2]

My point is that how to generate an integer by 31 bits, that valuable bit of it, is 1 no 0 . In fact RandomInteger[{a,b}] generate one integer between a and b. But RandomInteger[{2^30,2^31}] generate an integer by more than 31 bits , About 60 bits!!!

Also RandomInteger[{1,2^31}] generate an integer by about 60 bits ... !!!

Maybe This bug exist in mathematica6.0 and no in higher versions. Because I checked this problem by online mathematica in site http://www.wolframcloud.com/ and RandomInteger[{2^30,2^31}] generate an integer by 31 bits and Also RandomInteger[{1,2^31}] generate an integer by at most 31 bits .

Ok, your idea is good. Thank you for your help.

Have good times.