 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Make the least of these digits (Posted on 2003-04-13) You can use the digits 1,2,and 3 once only and any mathematical symbols you are aware of, but no symbol is to be used more than once. The challenge is to see if you can make the smallest positive number.

Special rules: You cannot use Euler's number or pi or infinity.

Special thanks to: Rhonda Wendel for Make the most of these digits and for the problem text which was slightly altered.

 No Solution Yet Submitted by Alan Rating: 3.0000 (10 votes) Comments: ( Back to comment list | You must be logged in to post comments.) That's it. This is the SMALLEST. I got it. | Comment 27 of 67 | Okay now I think I have finally obtained the smallest positive number which can be formed using the digits 1,2,and 3 once only and any mathematical symbols that we are aware of, and also no symbol being used more than once. Here is how I figured it out, though I do not know whether it is the smallest or not. So I might need help once again from Charlie or anyone who can help me with the calculations.

It has been observed from the previous comments that: 2^(-(31!)) is about 1/(10^(2.5*10^33)) while 3^(-(21!)) is about 1/(10^(2.4*10^19)), so the tiniest so far is 2^(-(31!)), and extended precision arithmetic tells us this is about 1.02192753424799x10^-2,475,321,084,412,797,072,581,101,485,711,599.

Now all I have to do is use a sine function because I know that sin(x) approaches 0 (zero) as x approaches 0 (zero).

Therefore the smallest positive number which can be formed using the digits 1,2,and 3 once only and any mathematical symbols that we are aware of, and also no symbol being used more than once is as given below:

sin [2^{-(31!)}].

That is, sin(x), where x = 2^(-(31!))

 Posted by Ravi Raja on 2003-04-15 23:07:54 Please log in:

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