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.

Charlie , or anybody else please could you help me out with this one? I have found out two different solutions but I cannot just figure it out which one of the two is smaller. Could you please perform the calculations for me as you do and let me know. Thank You. Here are the two different numbers (smallest positive) that I have obtained using the digits 1, 2 and 3 only once:

(a) [(3!)]^[-{(21!)}], which is equal to (1/6) raised to the power of 21 factorial. So that we have 1 in the numerator and 6 raised to the power (21!) in the denominator. And here is the next one.

(b) [(.1)]^[(2)^(3!)]!

Which is equal to (.1) raised to the power of (64!) (factorial of 64). Since (3!) = 6 and [(2)^(3!)] = 64 and therefore [(2)^(3!)]! = (64)!. This will give us a number which will contain (64!), that is, (factorial 64) zeros between the 'Decimal Point' and the digit 1(One).