Math game

[Crumb]

If you have some time to kill and you are not mathphobic here is a fun game to play. Try to find an equation using exactly four fours for as many numbers as you can. Use the following mathematical operations: + - ! ^ :sqroot: :x: :divide: (I think that covers it.)

Here are a few examples:

1 = (4+4) / (4+4)

2 = (4 / :sqroot:4) + 4 - 4
2 = 4/4 + 4/4

3 = (4 / :sqroot:4) + 4/4

4 = (4 / :sqroot:4) + (4 / :sqroot:4)
4 = :sqroot:4 + :sqroot:4 + 4 - 4

5 = 4 + :sqroot:4 - 4/4

:popcorn:

[Ymir's blood]

42

[Crumb]

42 = 4! + 4! - 4 - :sqroot:4

:ptht:

[Ymir's blood]

Well of course. 42=everything!

[justaman]

3,456,272.

:popcorn:

[Dingfod]

four +fore -for -phore = fuph (4 letters)

[Crumb]

3,456,272.
eek...You bastard. This is gonna keep me up tonight. :P

[Crumb]

This is as close as I could come:

4,194,280 = ((:sqroot:4)^(4!) / 4) - 4!

Hey at least I am within 738,008

[ceptimus]

6 = 4! / 4 + 4 - 4
7 = 4 + 4 - 4 / 4
8 = 4 + 4 + 4 - 4
9 = 4 + 4 + 4 / 4
10 = 4 + 4 + 4 / sqrt(4)

[slimshady2357]

11 =
12 = (4!/4)(4/(sqrt(4))
13 =
14 = 4 + 4 + 4 + sqrt(4)
15 = (4)(4) - 4/4
16 = (hmmmm, toughy :P) 4 + 4 + 4 + 4 and (4/4)(4)(4) and (sqrt(4))(sqrt(4))(sqrt(4))(sqrt(4)) etc.....
17 = (4)(4) + 4/4
18 = 4! - 4 - 4/sqrt(4)
19 = 4! - 4 - 4/4
20 = 4! - (sqrt(4))(4/sqrt(4))
21 = 4! - 4 + 4/4
22 = 4! - (4/4)(sqrt(4))
23 = 4! - (4/sqrt(4))/sqrt(4)

someone get 11 and 13, would ya :)

Adam

[slimshady2357]

This is as close as I could come:

4,194,280 = ((:sqroot:4)^(4!) / 4) - 4!

Hey at least I am within 738,008

Did a little more on this and I can get closer, lets see how close we can get :D

{sqrt(sqrt((4!)!))}{4} - 4! = 3550041.85 which is off by 93769.85

4! = 24
(4!)! = 620448401733239439360000
sqrt ((4!)!) = 787685471322.938290082864
sqrt(sqrt((4!)!)) = 887516.4625644324

Adam

[livius drusus]

Hey Adam: :sqroot:

[slimshady2357]

Hey Adam: :sqroot:

But that's more typing than sqrt() :P make it into :sq: and I might change over :tongue3:

Adam

[ceptimus]

How about

11 = 4 / .4 + 4 / 4

13 = 44 / 4 + sqrt(4)

Do you think .4 and/or 44 is cheating?

[slimshady2357]

How about

11 = 4 / .4 + 4 / 4

13 = 44 / 4 + sqrt(4)

Do you think .4 and/or 44 is cheating?

Well you'd have to ask Crumb really, but I think they're ingenious anyway :)

Adam

[ceptimus]

24 = 4 * 4 + 4 + 4
25 = 4.4(recurring) / (.4 * .4(recurring))

Still waiting for Crumb's ruling, but the recurring thing pushes the boundaries even further. :P

26 = 4 * 4 + 4 / .4
27 = (4 + 4 + 4) / .4(recurring)
28 = 44 - 4 * 4
29 = (4 / .4) / .4 + 4
30 = (4 + 4 + 4) / .4

[Crumb]

{sqrt(sqrt((4!)!))}{4} - 4! = 3550041.85

I only count three fours here.


Do you think .4 and/or 44 is cheating?
44 and .4 are different numbers. :ptht: But it was very creative. I don't think you can do 11 or 13 without cheating though.
4.4(recurring)
:no2:

25 = 4! + :sqroot:4 - 4/4

[ceptimus]

Even with your crazy rules :P we can get arbitrarily close to 11 and 13.

If we take, say, :sqroot::sqroot::sqroot::sqroot::sqroot::sqroot: 4 and repeat the square root function a few hundred times, then we can get as close to 1 as we like. I'll call this operation multiroot(4)

Then:

11 = 4 + 4 + 4 - multiroot(4)

13 = 4 + 4 + 4 + multiroot(4)

[Crumb]

I don't want "arbitrarily close". I want exact. Anyway, this is usually restricted to integers so :sqroot::sqroot:4 wouldn't be allowed. I don't think there is a 11 or 13 solution and I want it to stay that way damnit! :hmph:

[Ensign Steve]

Or what? You'll take your ball and go home?

[Crumb]

yes.

[slimshady2357]

{sqrt(sqrt((4!)!))}{4} - 4! = 3550041.85

I only count three fours here.

Excellent, then I can get even closer with:

{sqrt(sqrt((4!)!))}{4} - 4!^sqrt(4) = 3549469.85 which is only off by 93197.85! :)

Adam

[ceptimus]

I don't want "arbitrarily close". I want exact. Anyway, this is usually restricted to integers so :sqroot::sqroot:4 wouldn't be allowed. I don't think there is a 11 or 13 solution and I want it to stay that way damnit! :hmph:11 = (4! / :sqroot:4) - (4 / 4)

13 = (4! / :sqroot:4) + (4 / 4)
13 = (4! * :sqroot:4 + 4) / 4

Ha!

[slimshady2357]

I don't want "arbitrarily close". I want exact. Anyway, this is usually restricted to integers so :sqroot::sqroot:4 wouldn't be allowed. I don't think there is a 11 or 13 solution and I want it to stay that way damnit! :hmph:11 = (4! / :sqroot:4) - (4 / 4)

13 = (4! / :sqroot:4) + (4 / 4)
13 = (4! * :sqroot:4 + 4) / 4

Ha!

Lovely, thank you. :yup:

Adam

[Crumb]

Ha!
:qex:
:burns:

So, it looks like we have 1-25 & 42

26 = 4! + :sqroot:4 + :sqroot:4 - :sqroot:4
27 = 4! + 4 - (4/4)
28 = 4! + 4 + 4 - 4
29 = 4! + 4 + (4/4)
30 = 4! + 4 + 4 - :sqroot:4

[justaman]

This is as close as I could come:

4,194,280 = ((:sqroot:4)^(4!) / 4) - 4!

Hey at least I am within 738,008
:clap: You got closer than I thought you could...

Incidentally what does ! actually mean? I gather 4! is 24 but how cometh? :?

[Crumb]

! is called "factorial" it is used in combinatorial stuff.

3! = 3 * 2

4! = 4 * 3 * 2

5! = 5 * 4 * 3 * 2

etc...

[Skep]

You left out the good part, Crumb. 0!=1 :yup:

[justaman]

You left out the good part, Crumb. 0!=1 :yup:
Well that's just confusing...

[Skep]

From MathWorld (http://mathworld.wolfram.com/Factorial.html):

The special case 0! is defined to have value 0!=1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set).

[JoeP]

From MathWorld (http://mathworld.wolfram.com/Factorial.html):

The special case 0! is defined to have value 0!=1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set).
Justy, I presumed you're not confused any more? :muahaha:

[Shake]

Well then, of course
31 = 4! + 4 + (4/ :sqroot: 4)
32 = 4! + 4 + :sqroot: 4 + :sqroot: 4 OR 4! + ((4)(4)/:sqroot: 4)
34 = 4! + 4 + 4 + :sqroot: 4
36 = 4! + 4 + 4 + 4

[ceptimus]

Well then, of course
31 = 4! + 4 + (4/ :sqroot: 4)
32 = 4! + 4 + :sqroot: 4 + :sqroot: 4 OR 4! + ((4)(4)/:sqroot: 4)
34 = 4! + 4 + 4 + :sqroot: 4
36 = 4! + 4 + 4 + 4The first one here works out to 30, not 31

[seebs]

For any specified number, I can make a number larger than it.

4!!!!!!!! + 4 + 4 + 4

Add !'s to taste.

[justaman]

From MathWorld (http://mathworld.wolfram.com/Factorial.html):

The special case 0! is defined to have value 0!=1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set).
Justy, I presumed you're not confused any more? :muahaha:
No, now I'm just mad.

I have a severe philosophical objection to make about someone telling me that an empty set can be in any manner an 'arrangement'. It's like telling me non-existent Thunder-Couger-Falcon-Bird Sports Cars only come in one colour.

Maths is stupid. :fuming:

[ApostateAbe]

35 = 44/4 + 4!

I am still working on 31. It would be no problem if only I had another 4. It's driving me nuts.

[ceptimus]

The OP (Crumb) doesn't allow 44. I tried earlier to use 44 and .4 and recurring, but Crumb ruled them illegal. So sorry, you will have to try again on 35.

31 = (4! + 4) / 4 + 4!

37 = (:sqroot:4 + 4!) / :sqroot:4 + 4!

[Shake]

Well then, of course
31 = 4! + 4 + (4/ :sqroot: 4)
32 = 4! + 4 + :sqroot: 4 + :sqroot: 4 OR 4! + ((4)(4)/:sqroot: 4)
34 = 4! + 4 + 4 + :sqroot: 4
36 = 4! + 4 + 4 + 4The first one here works out to 30, not 31
:doh:

[Ensign Steve]

I have a severe philosophical objection to make about someone telling me that an empty set can be in any manner an 'arrangement'. It's like telling me non-existent Thunder-Couger-Falcon-Bird Sports Cars only come in one colour.

Duh. Red.

[slimshady2357]

The OP (Crumb) doesn't allow 44. I tried earlier to use 44 and .4 and recurring, but Crumb ruled them illegal. So sorry, you will have to try again on 35.

31 = (4! + 4) / 4 + 4!

37 = (:sqroot:4 + 4!) / :sqroot:4 + 4!

35 is right there, practically :)

35 = (4! - sqrt(4))/sqrt4 + 4!
36 = (4!/4)(4!/4)
37 = ceptimus' answer above
38 = 4! + 4!/sqrt(4) + sqrt(4)
39 =
40 = 4! + 4! - 4 - 4
41 =
42 = 4! + 4! - 4 - sqrt(4)
43 =
44 = 4! + 4! - sqrt(4) - sqrt(4)
45 =
46 = 4! + 4! - 4 + sqrt(4)
47 = 4! + 4! - 4/4
48 = 4! + 4! + 4 - 4
49 = 4! + 4! + 4/4
50 = 4! + 4! + 4 - sqrt(4)

So can anyone get 39, 41, 43, or 45?

Adam

[ceptimus]

I don't think those are possible within Crumb's rules. Maybe we'll find one of them, but I think at least one is impossible.

If we can use 44 and .4 then I have solutions to them all. Here are a few more that are legal under the rules.

58 = (4⁴ - 4!) / 4

63 = (4⁴ - 4) / 4

64 = :sqroot::sqroot::sqroot: (4^4!) + 4 - 4

65 = (4⁴ + 4) / 4

68 = 4 x 4 x 4 + 4

70 = (4⁴ + 4!) / 4

78 = 4(4! - 4) - :sqroot:4

80 = 4(4 X 4 + 4)

81 = (4/4 - 4)⁴

The 64 solution, that only really uses two of the fours, is a bit of a shocker, and provides the basis for quite a few other solutions...

I like the 81 solution. :)

[Shake]

78 = 4(4! - 4) - :sqroot:4
Following this lead, we have:

76 = 4(4! - 4) - 4
82 = 4(4! - 4) + :sqroot:4
84 = 4(4! - 4) + 4
I like the 81 solution. :)
Me too! That's quite clever!

[ceptimus]

52 = :sqroot::sqroot::sqroot: (4^4!) - 4!/:sqroot:4

56 = :sqroot::sqroot::sqroot: (4^4!) - (4 + 4)

62 = :sqroot::sqroot::sqroot: (4^4!) - 4 / :sqroot:4

66 = :sqroot::sqroot::sqroot: (4^4!) + 4 / :sqroot:4

72 = :sqroot::sqroot::sqroot: (4^4!) + 4 + 4

[livius drusus]

I'm not good enough to participate, but I just want to say that I love this thread and it's a genuine pleasure seeing what y'all come up with. :thumbup:

[John Carter]

From MathWorld (http://mathworld.wolfram.com/Factorial.html):

The special case 0! is defined to have value 0!=1, consistent with the combinatorial interpretation of there being exactly one way to arrange zero objects (i.e., there is a single permutation of zero elements, namely the empty set).
Justy, I presumed you're not confused any more? :muahaha:
No, now I'm just mad.

I have a severe philosophical objection to make about someone telling me that an empty set can be in any manner an 'arrangement'. It's like telling me non-existent Thunder-Couger-Falcon-Bird Sports Cars only come in one colour.

Maths is stupid. :fuming:

First, Mathworld is not a very good resource. It is full of small errors (IOW, there are nits that can be picked with many of the pages at that site). They can't even get the difference between the Well-Ordering Principle and the Well-Ordering Theorem straight!

You have misinterpreted what is being said here. Read that statement again. In no way does it way that the empty set is in any way an arrangement. It says that there is no way to arrange zero elements, so the set of possible arrangements is the empty set and since the empty set is itself a valid set, there is one set of possible arrangements.

The factorial is defined recursively, in general, for a natural number n, n! = n * (n-1)!. This means we must have 0! be defined as something, and 1 makes everything "work out". IOW: 0! = 1 lets us say 1! = 1 * (1-1)! = 1 * 0! = 1 *1 =1. consistent with our definition, while if we said 0! = 0, then for any n, n! =0 which is a rather useless operation..

[Dragar]

The factorial is defined recursively, in general, for a natural number n, n! = n * (n-1)!

Ah-ha!

Thanks, John, that just made it 'click' for me.

[JoeP]

The factorial is defined recursively, in general, for a natural number n, n! = n * (n-1)!. This means we must have 0! be defined as something, and 1 makes everything "work out". IOW: 0! = 1 lets us say 1! = 1 * (1-1)! = 1 * 0! = 1 *1 =1. consistent with our definition, while if we said 0! = 0, then for any n, n! =0 which is a rather useless operation..
I agree, except why stop at 0? (-1)! "must be defined as something", and 0! = 0 * (-1)! Then either 0! = 0, and we still don't know what (-1)! is, or (-1)! is infinite but the recursion doesn't tell what 0! is. (The latter applies, in any reasonable sense, I assume.)

This is kind of an amble, but in writing I've hit on a rephrasing of your words: If 0! is to be defined as anything, it must be 1, to make the recursion 1! = 1 * (1-1)! = 1 * 0! = 1 *1 = 1 work.

[Skep]

Those who are interested in the factorial function may want to read up on the gamma function which extends the factorial function to real and complex numbers. The gamma function also has interesting connections to the Riemann Zeta Function. Perhaps John Carter would be kind enough to provide a link to a reliable resource.

[JoeP]

I'll happily post an unreliable resource (http://en.wikipedia.org/wiki/Gamma_function), if only to confirm the implication in my previous post that (-1)! is infinite (strictly, undefined).

[Zephyrus]

45 = (4 + sqrt(4))!/(4^sqrt(4))

[Zephyrus]

while further messing around, i came across this nifty guy that should work, though i don't have a calculator to test it on and my random trig identity knowledge is a bit rusty:

sec(arctan(sqrt(4 + sqrt(4))))^(sqrt(4)) - 4 = 45

still working on 39, 41, 43...

[ceptimus]

Crumb didn't allow the use of trig functions but that sec and tan business is a neat way of getting a 7. :) It uses the right angled triangle with sides 1, sqrt(6), and sqrt(7). If trig functions were to be allowed, you could use similar tricks to get (for example) 17 using the triangle with sides 1, sqrt(4 x 4) and sqrt(17).

ETA: Welcome to the forum Zephyrus.

[Crumb]

Welcome to the forum zephyrus. The trig functions are very cool. :1thumbup: