This is a real story. A long long time ago, based on the "Four fours" problem, I wondered if I could do the same using exactly five 3īs, the restrictions a bit tighter: I could use only the 4 basic math operations, exponentiation, factorial, and all parentheses I may need. Besides, I didnīt disallow me to join two "3īs" to make "33".
Using this, and only this, I succeeded in writing expressions for all integers from 0 to 100.
To narrow your work, since a great number of integers can be easily obtained, can you find expressions for 47, 50, 56, 58, 64, 70, 71, 73, 74, 76, 77, 85, 88, 94, and 95?
47 : 3! * 3! + 33/3
50 : (3! + 3!/3)!/(3!)!  3!
56 : 3! * 3 * 3 + 3!/3
58 : (3 + 3/3)
^{3}  3!
64 : (3 + (3 + 3)/3!)
^{3}70 : 3!
^{3}/3  3!/3
71 : 3!
^{3}/3  3/3
73 : 3!
^{3}/3 + 3/3
74 : 3!
^{3}/3 + 3!/3
76 : (3!)!  3! * 3!)/(3 * 3)
77 : (3!)!/(3 + 3 + 3)  3
85 : (3!)!/3 + 3!)/3 + 3
88 : (3!)!/3 + 3!)/3 + 3!
94 : ((3!)! + 3!)/3!  3
^{3}95: 3!
^{3}  ((3!)! + 3!)/3!
I have not (yet) found solutions for 65, 68, and 91 under the tighter restrictions.
Using the decimal (.3), which not explicitly prohibited in the problem (though mentioned as not allowed in the queue comments), I provide here solutions to those unachieved under the tighter restrictions:
65 : (3! * 3! + 3)/(.3 + 3)
68 : ((3!)! * .3  3!  3!)/3
91 : 3
^{3}/.3 + 3/3
Edited on July 8, 2008, 1:36 am

Posted by Dej Mar
on 20080707 19:47:45 