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.)?

since ( n-.5)/.5=2*n-1

and ( n+.5)/.5=2*n+1

2 new numbers can be generated from any uq(3) by using 2 additional fives

and a lot more from any uq(2)

e.g. 480 ===> 959,961; 555===>1111,1109

720=====> 1439,1441(and to each of those two one can add or subtract any uq(1) ,,,, 1319=1439-5! etc.

by the same token any n=uq(2) can generate 2*n-5 and and 2*n+5 using 2*( n-.5*5) and 2*( n+.5*5)

600===>1205 , 1195

substitute 5.5 for .5*5 - one gets 2*n-11 and 2*n+11

one more trick ,starting with n=uq(1)

(n+.5)^2-.25 is an integer ,so is (n-.5)^2-.25

.5 root(!5-.5)-.5*.5=1892

.5 root(!5+.5)-.5*.5=1980

now imagine someone asking you 3 months ago to express 1892

using 5 fives.....

I leave it to you to find out if the above advices bring any

new numbers (now we are looking for three digits numbers-

who knows where to stop??)

good luck, Jim (a quote from "Mission Impossible"