Answer in 5 . 4 . 3 . 2 . 1 .....

[science_man_88]

using straight addition we can make all of 1,3,6,10,15 aka the triangular numbers up to 15 noting that the squares are sums of odd numbers we can get the first 3 squares plus 0,2,4,or 6 ( some of these overlap the triangular numbers of course) so we can produce 1,4,9,3,6,11,5,8,13,7,10,15 all through addition only.we can't subtract 4 and 5 both at the same time since (4+5)>(1+2+3) if we form fractions ( using parentheses and division) then we know certain arrangements don't turn out integers so those combinations are out including things like (1+3+5)/(2+4) in fact any sum on the top need be even I think as even/odd odd/odd and even/even are possible in theory but I think any odd/odd ones don't turn out integer in this case. (1+2+3+4)/5 =2 works though. there are limits on each operation but I haven't really thought much deeper. never mind I forgot the order restriction. there's still the limits to consider though.

[VBCurtis]

One more cheater:
39 = 5 * 4! / 3 - 2 + 1

[xilman]

Originally Posted by KangJ

1=5/(4*3/2-1), 1=(5-4+3)/2-1
...
49=5*(4*3-2*1)

I think 39 is only the impossible number without cheating

I feel it unfair that the question excludes 0 from the list to be created. Creating negative numbers is, of course, trivial, given an expression for the corresponding positive. In a spirit of fairness, I offer 0 = (5-4-3+2)*1 as a representative of several others.

Now that everyone has had their fun, I'm tempted to write a Perl script which evaluates all possible binary trees of the form (tree operator tree) where the terminal nodes are taken from the set {5,4,3,2,1} and operator from the set {+, -, *, /} subject to the constraint that the resulting expression has its leaves in descending order.

Clearly, the maximum value which can be created without cheating is 120, the smallest is -120; some values are not integers, one of which is 5/(4*3*2*1).

[science_man_88]

Originally Posted by xilman

Clearly, the maximum value which can be created without cheating is 120, the smallest is -120; some values are not integers, one of which is 5/(4*3*2*1).

nope 121= 5*4*3*2+1 = 120+1

[xilman]

Originally Posted by science_man_88

nope 121= 5*4*3*2+1 = 120+1

Well spotted!

[R. Gerbicz]

I've found all possible integers with a simple c code (note that here I have not allowed unary operations), as you can see you really can't reach 39.
(in some lines you can delete some parentheses to get the same number)

-31=5-(4*(3*(2+1)))
-25=5*(4-(3*(2+1)))
-23=5-(4*(3*2+1))
-21=(5-(4*3))*(2+1)
-20=5-(4*(3*2)+1)
-19=5-(4*(3+2+1))
-18=5-(4*(3*2)-1)
-16=5-((4+3)*(2+1))
-15=5-(4*(3+2*1))
-14=5-(4*(3+2)-1)
-13=(5-(4*3))*2+1
-11=5-(4*(3+2-1))
-10=5*(4-(3+2+1))
-9=5-((4+3)*(2*1))
-8=5-(4+3*(2+1))
-7=5-(4*(3*(2-1)))
-6=5-(4+3*2+1)
-5=5-(4+3+2+1)
-4=5-(4+3+2*1)
-3=5-(4+3+2-1)
-2=5-(4+3*(2-1))
-1=5/(4-(3*(2+1)))
0=5*(4*(3-(2+1)))
1=5-(4+3-(2+1))
2=5-(4-(3/(2+1)))
3=5+4-(3+2+1)
4=5-(4/(3+2-1))
5=5+4*(3-(2+1))
6=5+4/(3+2-1)
7=5-(4-(3+2+1))
8=5+4-(3/(2+1))
9=5+4+3-(2+1)
10=5-(4-(3*(2+1)))
11=5+4+3-(2-1)
12=5+4+3*(2-1)
13=5+4+3+2-1
14=5+4+3+2*1
15=5+4+3+2+1
16=5+4+3*2+1
17=5+4*(3*(2-1))
18=5+4+3*(2+1)
19=5+(4+3)*(2*1)
20=5*(4+3-(2+1))
21=5+4*(3+2-1)
22=5*4+3-(2-1)
23=5*4+3*(2-1)
24=5+4*(3+2)-1
25=5*(4+3/(2+1))
26=5+(4+3)*(2+1)
27=(5+4)*(3*(2-1))
28=5+4*(3*2)-1
29=5+4*(3+2+1)
30=5*(4+3-(2-1))
31=5*(4*(3/2))+1
32=5*(4+3)-(2+1)
33=5+4*(3*2+1)
34=5*(4+3)-(2-1)
35=5*(4+3*(2-1))
36=(5+4)*(3+2-1)
37=5*(4+3)+2*1
38=5*(4+3)+2+1
40=5*(4+3+2-1)
41=5+4*(3*(2+1))
44=5*(4+3+2)-1
45=5*(4+3+2*1)
46=(5*4+3)*(2*1)
47=(5*4+3)*2+1
49=5*(4+3*2)-1
50=5*(4+3+2+1)
51=(5+4*3)*(2+1)
53=(5+4)*(3*2)-1
54=(5+4)*(3+2+1)
55=5*(4+3*2+1)
57=5*(4*3)-(2+1)
58=5*(4*3)-(2*1)
59=5*(4*3)-(2-1)
60=5*(4*(3*(2-1)))
61=5*(4*3)+2-1
62=5*(4*3)+2*1
63=(5+4)*(3*2+1)
65=5*(4+3*(2+1))
69=(5*4+3)*(2+1)
70=5*((4+3)*(2*1))
71=5*((4+3)*2)+1
75=5*(4*3+2+1)
80=5*(4*(3+2-1))
81=(5+4)*(3*(2+1))
95=5*(4*(3+2)-1)
99=5*(4*(3+2))-1
100=5*(4*(3+2*1))
101=5*(4*(3+2))+1
105=5*((4+3)*(2+1))
115=5*(4*(3*2)-1)
119=5*(4*(3*2))-1
120=5*(4*(3+2+1))
121=5*(4*(3*2))+1
125=5*(4*(3*2)+1)
140=5*(4*(3*2+1))
180=5*(4*(3*(2+1)))

[Batalov]

I can represent a (reasonably) large prime with these digits and only one sign symbol :-)
54³² + 1 = 27327525884414205519790497974303154461449992065060438017

[rogue]

Originally Posted by R. Gerbicz

I've found all possible integers with a simple c code (note that here I have not allowed unary operations), as you can see you really can't reach 39.
(in some lines you can delete some parentheses to get the same number)

Now you just need to do it with the three 6's puzzle...

[R. Gerbicz]

Originally Posted by rogue

Now you just need to do it with the three 6's puzzle...

Not following every topics, what/where is that puzzle?

[Uncwilly]

Originally Posted by KangJ

39=54/3+21
I think 39 is only the impossible number without cheating

I think that if you read the rules carefully, you have not cheated.

Originally Posted by petrw1

Another way to get 39 with bending another rule:
39 = 5 X ( (3 X 2) + 1) + 4

That is rule breaking.

Originally Posted by bsquared

Another cheat :
39 = (5 * (4 + 3) + 2 + 1)++

Originally Posted by VBCurtis

One more cheater:
39 = 5 * 4! / 3 - 2 + 1

Since the rules (as provided in the original link calls 5, 4, 3, 2, & 1 "digits" and not numbers, I think KangJ is ok.
Similarly:
39 = -5 + 43 + 2 - 1

[rogue]

Originally Posted by R. Gerbicz

Not following every topics, what/where is that puzzle?

Maybe that was wrong. I'm trying to find it. Possibly it was six sixes not three sixes. I'm sure that someone else here will remember.