A mathematician who was exceedingly fond of the number five set to work trying to express as many consecutive integers using no numerals besides '5', and only up to five of them. She allowed herself to use any standard mathematical notation she knew, as long as it didn't contain any numerals. For example, she could use the symbol for 'square root', but not 'cube root' (because it contains a '3'). She determined that the highest consecutive integer she could express this way was 36. Her last few calculations were as follows:
 31 = 5*5 + 5 + (5/5)
 32 = 55*.5 + 5  .5
 33 = (55 + 5) * .55
 34 = 5!/5 + 5/.5
 35 = (5 + (5+5)/5) * 5
 36 = 5*5 + 55/5
 37 = ?
Was she correct in thinking 36 was the highest consecutive integer she could express this way? Can you express 37 using only up to
five 5's?
Note: The intention here is to find an exact expression, so rounding expressions like [] "greatest integer" are not allowed.
Note: Can you do it without using letters of any kind (x, log, lim, sum, etc.)?
Using superfactorials (n$), primorials (n#) and mulitfactorials, in addition to the double factorials (n!!), and subfactorials (!n) in many of the early numbers in the list can reduce the expressions by one or more fives....Following is a small list of some that can be reduced (I've also tried to reduce the notations following any reduction of the number of 5's used):
2 = 5#/5!!
3 = 5!!/5
4 = 5!/5#
6 = 5#/5
8 = 5!/5!!
10 = 5!!!
12 = (5!!/5)$
13 = 5!/5!! + 5
14 = !5  5#
15 = 5!!
16 = (5!/5!!)!!!!!!
17 = 5!! + 5#/5!!
18 = (5#/5)!!!
20 = (5!!!)!!!!!!!!
21 = (SQRT(!5 + 5))!!!!
22 = .5 * !5
26 = 5#  5!/5#
27 = 5!!/.5
28 = (SQRT(!5 + 5))!!!
29 = !5  5!!
30 = 5#
31 = 5# + 5/5
33 = (5#/5)!!! + 5!!
34 = !5  5!!!
35 = 5# + 5
36 = (5/.5)!!!!!
Edited on May 11, 2006, 6:01 am

Posted by Dej Mar
on 20060216 05:01:50 