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Using only the numbers 1 to 4, with each being used as a single digit once in each case, can the numbers 1 through 40 be created using only these numbers plus mathematical symbols? e.g. 21 = (1+2).(3+4)

Note: (1) All numbers 1-4 must be used in each case, and each must be distinct (using them as numbers, not digits, so e.g. 34-21 is not allowed), and (2) Warning - the answer may be no! (I don't know if all are possible myself, yet, though I know most are).

Can 4 be used as a 2 simply by including (√4) in an equation?

or is it only +, -, x, /, ^ that can be used?

1 = (1+4)/(2+3)
2 = 1+2+3-4
3 = (3+4-1)/2
4 = ((1+3)x2)-4
5 = ((1+2)x3)-4
6 = ((3x4)/2)x1
7 = ((4+1)x2)-3
8 = 2-1+3+4
9 = (3x4)-(1+2)
10 = ((3x4)-2)x1
11!back later

DRT wrote:1 = (1+4)/(2+3)
2 = 1+2+3-4
3 = (3+4-1)/2
4 = ((1+3)x2)-4
5 = ((1+2)x3)-4
6 = ((3x4)/2)x1
7 = ((4+1)x2)-3
8 = 2-1+3+4
9 = (3x4)-(1+2)
10 = ((3x4)-2)x1
11!back later

11 = 3x4-2+1
12...

I started working back from 40.

40. ((3^4)-1)/2

39. Now stuck!

RAYC wrote:Can 4 be used as a 2 simply by including (√4) in an equation?
or is it only +, -, x, /, ^ that can be used?

The √ symbol is acceptable (though possibly not much help).

Depressingly this "minor" diversion seems more exciting than knuckling down to work, and has become a major diversion!

12.(4x3)/(2-1)
13. 4x3+2-1
14. 4x3x1+2
15. 4x3+2+1
16. 4x(3+(2-1))
17. (4+1)x3+2
18. 4x(3+1)+2
19. 4x(3+2)-1
20. (2^4)+3+1
21. (4+2+1)x4
22. ((4+1)^2)-3
23. ((2+1)^3)-4
24. √((3+2)^4)-1
25. √((3+2)^4)x1
26. √((3+2)^4)+1
27. ((2+1)^4)/3
28. ((4+1)^2)+3
29. (2^(4+1))-3
30. 2x3x(4+1)
31. ((2+1)^4)+4
32. (2^4)x(3-1)
33. ((4^3)/2)+1

DRT wrote:1 = (1+4)/(2+3)
2 = 1+2+3-4
3 = (3+4-1)/2
4 = ((1+3)x2)-4
5 = ((1+2)x3)-4
6 = ((3x4)/2)x1
7 = ((4+1)x2)-3
8 = 2-1+3+4
9 = (3x4)-(1+2)
10 = ((3x4)-2)x1

AHB wrote:11 = 3x4-2+1

12 = (2x4)+1+3
13 = (3x4)+2-1
14 = ((3x4)+2)x1
15 = (3x4)+2+1
16 = (2+3-1)x4
17 = (4²+3-2)x1

Pause: is the use of the mathematical symbol ² permitted?

We need to find a way to avoid two or more people wasting time simultaneously

DRT wrote:Pause: is the use of the mathematical symbol ² permitted?

It feels wrong to me, but if you are allowed to use √ then it is hard to argue against it i suppose!!

39 then becomes very easy indeed... (3²)x4+2+1

Phil to adjudicate - i am happy to remove the use of √

DRT wrote:

DRT wrote:1 = (1+4)/(2+3)
2 = 1+2+3-4
3 = (3+4-1)/2
4 = ((1+3)x2)-4
5 = ((1+2)x3)-4
6 = ((3x4)/2)x1
7 = ((4+1)x2)-3
8 = 2-1+3+4
9 = (3x4)-(1+2)
10 = ((3x4)-2)x1

AHB wrote:11 = 3x4-2+1

12 = (2x4)+1+3
13 = (3x4)+2-1
14 = ((3x4)+2)x1
15 = (3x4)+2+1
16 = (2+3-1)x4
17 = (4²+3-2)x1

Pause: is the use of the mathematical symbol ² permitted?

That would count as use of the digit 2. In the same manner, 2√ or 3√ could be used for square root or cube root, thereby using the digit 2 or 3 as well as a 'symbol'.

√ by itself is the symbol for square root, no?

No need to put a 2 in front...

But happy to remove use.

RAYC wrote:√ by itself is the symbol for square root, no?

No need to put a 2 in front...

Indeed, though optionally you can put any digit in front to represent the n-th root - which in this case counts as using the digit, just as using the digit within a power would whether you use the form 3² or 3^2. Each digit must be used individually (i.e. not compounds such as 13) and if you can see the digit in the written expression, then it's been used.

OK. So my solution to 17 is void, but replaced by Rob's.

Rob's solution to 31 uses two 4s and no 3.

DRT wrote:Rob's solution to 31 uses two 4s and no 3.

It also makes 85, not 31.

It should be ((2+1)^3)+4.

DRT wrote:

DRT wrote:Rob's solution to 31 uses two 4s and no 3.

It also makes 85, not 31.

It should be ((2+1)^3)+4.

yes - mis-typed.

24-26 also need to be re-arranged in light of the conversation above re: use of square root.

I have 31,32,35,36,37 and 40 done...but 34 and 39 are troubling me!

Y'all are making some of these way more complicated than necessary...

10: 1+2+3+4
23: 2*3*4-1
24: 1*2*3*4
25: 2*3*4+1

I can get 39, but I can't express it with a keyboard.

Use sigma to express:

Sum(1..(4*2))+3

And following from that, 34 becomes:

Sum(3..(4*2))+1

Here we go. Still not entirely sure what the "proper" notation is with a keyboard, but at least I can express sigma now.

39: 1∑(4*2)+3
34: 3∑(4*2)+1

Or perhaps...

39: ∑(1..4*2)+3
34: ∑(3..4*2)+1

Glenn E. wrote:Here we go. Still not entirely sure what the "proper" notation is with a keyboard, but at least I can express sigma now.

39: 1∑(4*2)+3
34: 3∑(4*2)+1

Or perhaps...

39: ∑(1..4*2)+3
34: ∑(3..4*2)+1

Yes...can't say i would have got there myself!

Glenn E. wrote:Y'all are making some of these way more complicated than necessary...

Like the best swiss watches...!

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.

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!

Perhaps this discussion has reached a stage where is should be posted here?

If anyone feels brave enough to step into that particular world of geekism

From that place, a fabulous example of an entire discussion which neatly demonstrates the tendency of mathematicians to be brief and to the point!

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.