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Make the least of these digits (Posted on 2003-04-13) Difficulty: 3 of 5
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)

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Solution re: That's it. This is the SMALLEST. I got it. | Comment 60 of 67 |
(In reply to That's it. This is the SMALLEST. I got it. by Ravi Raja)

But, .1^(32!) = 1/(10^(32!)), or 1/(10^263130836933693530167218012160000000).
sin(1/(10^263130836933693530167218012160000000)) is even smaller.

Although these numbers are inconceivably large, and small, using primorial, we can get: sin [.1^(32!)#] where # is the primorial function. For those of you who don't know, primorial is the product of all primes less than or equal to x.

Now, take the sqrt(x) equals:
Use this as an estimate of the average product of the last prime and first prime, second to last and second,etc.

Even with only 4 primes, (.1^(32!)#) is already at least that huge number. So, the actual value of sin(.1^((32!)#)), will be so close to 0 that it would be impossible to ever tell the difference.

Just to show how rapidly the values decrease, look at the following:

.1^((2!)#) = 1/(10^2)
.1^((3!)#) = 1/(10^30)
.1^(((3+2)!)#) = 1/(10^(3.161e46))
.1^(((3*2)!)#) = 1/(10^(9.924e297)) which is about (100.33)/(googolplex^3) !!!!!!!!!

So, sin{.1^([32!]#)} is the smallest value that I even want to try and think about.

Edited on March 17, 2006, 8:54 pm
  Posted by Justin on 2006-03-17 20:49:09

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