ThePortForum.com

RAYC wrote:

PhilW wrote:A good show! Am slightly unsure on the sigma notation validity?
I would normally have expected limits above and below, such as
4*2
∑(n) +3
n=1

although this needs additional algebra letters, which is perhaps borderline on being allowed...
If ∑(1..4*2) is acceptable to mean the sum of the list of numbers represented then I would have to allow it.

I'll share my alternatives:
34 = 4! +3^2 +1
39 = (3!)^2 +4 -1

Use of factorial makes at least 1-52 possible, not sure how much higher.

Not to say that these aren't impressive, but summation and factorial operators are short-form definitions for much longer calculations (at least according to my very shaky understanding). So if square and square roots symbols can't be used without counting those as use of the number 2, i'm not convinced that these formulas can be used without counting all the numbers they involve when set out in long form!

Yes, but Phil already ruled that √n, which is itself the short form of 2√n, is valid. 4! does in fact represent 4*3*2*1 but it is a standard mathematical operator (symbol) and therefore should be allowed. As ruled, you must count the use of a digit if it is visible in the notation, thus making 4! a valid way to use the digit 4 and only the digit 4.

My concern with ∑ was that it might not be considered a standard mathematical symbol because it is a Greek letter. However Phil has reminded me that its standard use does involve further algebraic notation so I don't think it should be valid. Then again, due to the prevalence of spreadsheets it might be that ∑(1..4*2) is now valid notation, so I really don't know.

This also begs the question, what is the largest number you can represent using standard mathematical notation and the digits 1, 2, 3, and 4 under the same rules that Phil has laid out?

2^3!^(4+1)! = 64^120 = 5.515652e216

or better yet...

3!^((4*(2+1))!) = 6^479001600 which causes my calculator to give up and simply respond with ∞.

Glenn E. wrote:This also begs the question, what is the largest number you can represent using standard mathematical notation and the digits 1, 2, 3, and 4 under the same rules that Phil has laid out?

2^3!^(4+1)! = 64^120 = 5.515652e216

or better yet...

3!^((4*(2+1))!) = 6^479001600 which causes my calculator to give up and simply respond with ∞.

I don't think there is a limit; you could keep on adding more brackets and factorial signs:
((((((2^(3^(4+1)))!)!)!)!)!)! etc

PhilW wrote:I don't think there is a limit; you could keep on adding more brackets and factorial signs

Hmm... good point. So I guess it's not a very interesting question.

Oh my, that's a lot of numbers, good thing it's #PortDay. I think I'll start drinking a little early...Like now

Just remembered this thread. I can do 1 to 84, except 76.

1 = ((1^3)^(4/2))
2 = ((1+2)+(3-4))
3 = ((1*2)-(3-4))
4 = ((1+2)-(3-4))
5 = ((1*3)-(2-4))
6 = ((1-2)+(3+4))
7 = ((1^2)*(3+4))
8 = ((1^2)+(3+4))
9 = ((1*2)+(3+4))
10 = ((1+2)+(3+4))
11 = ((1-2)+(3*4))
12 = ((1^2)*(3*4))
13 = ((1^2)+(3*4))
14 = ((1*2)+(3*4))
15 = ((1+2)+(3*4))
16 = ((1^3)*(2^4))
17 = ((1^3)+(2^4))
18 = ((1*3)*(2+4))
19 = ((1*3)+(2^4))
20 = ((1+3)+(2^4))
21 = ((1+2)*(3+4))
22 = (2*((3*4)-1))
23 = ((2*(3*4))-1)
24 = ((1*2)*(3*4))
25 = ((1+4)*(2+3))
26 = ((1+(3*4))*2)
27 = ((3^2)*(4-1))
28 = ((1+(2*3))*4)
29 = ((2^(1+4))-3)
30 = ((1+4)*(2*3))
31 = ((3^(1+2))+4)
32 = ((1/2)*(4^3))
33 = ((4*(2^3))+1)
34 = ((1+(4!))+(3^2))
35 = ((2^(1+4))+3)
36 = ((1+2)*(3*4))
37 = ((4*(3^2))+1)
38 = ((2^(1+4))+(3!))
39 = (((3!)^2)-(1-4))
40 = ((1+4)*(2^3))
41 = ((1+(3^4))/2)
42 = ((1*2)*((4!)-3))
43 = ((2*((4!)-1))-3)
44 = ((1-3)*(2-(4!)))
45 = ((1+4)*(3^2))
46 = ((1-3)+(2*(4!)))
47 = ((3*(2^4))-1)
48 = ((1*3)*(2^4))
49 = ((3+4)^(1*2))
50 = (1+((3+4)^2))
51 = ((1+(2^4))*3)
52 = ((1+3)+(2*(4!)))
53 = ((2*(1+(4!)))+3)
54 = ((3^(4-1))*2)
55 = ((2*(3+(4!)))+1)
56 = ((1+(3!))*(2*4))
57 = (4+1)!/2-3
58 = ((4^(1+2))-(3!))
59 = ((2^(3!))-(1+4))
60 = ((1+4)*(2*(3!)))
61 = ((4^3)-(1+2))
62 = ((4^3)-(1*2))
63 = ((1-2)+(4^3))
64 = ((1^2)*(4^3))
65 = ((1^2)+(4^3))
66 = ((1*2)+(4^3))
67 = ((1+2)+(4^3))
68 = ((1*4)+(2^(3!)))
69 = ((1+4)+(2^(3!)))
70 = ((3*(4!))-(1*2))
71 = ((1-2)+(3*(4!)))
72 = ((1+2)*((3!)*4))
73 = ((1^2)+(3*(4!)))
74 = ((1*2)+(3*(4!)))
75 = (3*((1+4)^2))
76 = ?
77 = ((3*(1+(4!)))+2)
78 = ((3^4)-(1+2))
79 = ((3^4)-(1*2))
80 = ((1-2)+(3^4))
81 = ((1^2)*(3^4))
82 = ((1^2)+(3^4))
83 = ((1*2)+(3^4))
84 = ((1+2)+(3^4))

jdaw1 wrote:Just remembered this thread. I can do 1 to 84, except 76.

Well played. My only solution for 76 would be
76 = (∑(1..(3*4)))-2
provided that such use of ∑(1..n) is valid to mean the sum of values 1 to n without the full algebraic form required. Without the ∑ I do not have a solution for 76 either.