c4countdown

Posted: **Sun Jun 14, 2009 11:43 am**

Liam Tiernan wrote:Next one..........not so easy.

No problem if we use recurring decimals.

                        .
(sqrt(4)xsqrt(4))! + 4/.4 = 33

Posted: **Sun Jun 14, 2009 11:59 am**

4!+4+4+sqrt4=34
4!+(44/4)=35
4!+4+4+4=36
44-(4/.4)=37
44-4-sqrt4=38
44-sqrt4-sqrt4=40
44-4+(4/4)=41
44-4=sqrt4=42
44-(4/4)=43
44-4=4=44
44=(4/4)=45
44+4-sqrt4=46
4!+4!-(4/4)=47
4!+4!=4-4=48
4!+4!+(4/4)=49
4!+4!+(4/sqrt4)=50

Posted: **Sun Jun 14, 2009 12:29 pm**

Liam Tiernan wrote: 44-(4/.4)=37

No. This makes 34.

As I see it, the missing ones are now:
37, 39, and 51 onwards.

Posted: **Sun Jun 14, 2009 12:49 pm**

Howard Somerset wrote:
Liam Tiernan wrote: 44-(4/.4)=37
No. This makes 34.

As I see it, the missing ones are now:
37, 39, and 51 onwards.

Oops. Still can't see 37.
And hadn't noticed that I'd skipped 39.

Posted: **Sun Jun 14, 2009 1:06 pm**

Howard Somerset wrote:As I see it, the missing ones are now:
37, 39, and 51 onwards.

4*(!4)+(4/4)=37

44-(!4)+4=39

44+(!4)-Sqr4=51

Posted: **Sun Jun 14, 2009 1:23 pm**

16 = 4 + 4 + 4 + 4.

Posted: **Sun Jun 14, 2009 1:49 pm**

Daniel O'Dowd wrote:
(!4)

What function is this? Looks pretty helpful.

Posted: **Sun Jun 14, 2009 2:08 pm**

Innis Carson wrote:
Daniel O'Dowd wrote:
(!4)
What function is this? Looks pretty helpful.

I think he means .4 recurring[uhttp://en.wikipedia.org/wiki/Recurring_decimal#Notationrl][/url]
in which case 4!+4+(4/.4recurring)=37

Posted: **Sun Jun 14, 2009 2:54 pm**

Innis Carson wrote:
Daniel O'Dowd wrote:
(!4)
What function is this? Looks pretty helpful.

!N is the subfactorial, which gives the number of arrangements of N objects, none of which are in their original positions: it is the nearest integer to N!/e.

Posted: **Sun Jun 14, 2009 3:01 pm**

Liam Tiernan wrote:
Innis Carson wrote:
Daniel O'Dowd wrote:
(!4)
What function is this? Looks pretty helpful.
I think he means .4 recurring[uhttp://en.wikipedia.org/wiki/Recurring_decimal#Notationrl][/url]
in which case 4!+4+(4/.4recurring)=37

I most certainly do not!

Decimals aren't at all needed. Subfactorial is related to binomial expression, like 5C3 and Pascal's Triangle. =)

Continuing:

(4!-Sq4+4)x(Sq4)=52

Posted: **Sun Jun 14, 2009 3:06 pm**

4! + 4! + !4 - 4 = 53

Posted: **Sun Jun 14, 2009 3:37 pm**

So we're not allowed the gamma function?
When I was set this at secondary school we were allowed:
Bidmas, factorial, decimal point.
I see we've taken indices out of the equation too. I'm going to do my best at being a purist and stick to just those.

Posted: **Sun Jun 14, 2009 3:39 pm**

4! + 4! + 4 + sqrt(4) = 54

Posted: **Sun Jun 14, 2009 3:46 pm**

You can do 33 without recurring decimals:
4!+4+(sqrt(4)/.4)

Posted: **Sun Jun 14, 2009 3:51 pm**

37 and 39 without the subfactorial:
4!+(4!+sqrt 4)/sqrt 4 = 37
44-(sqrt(4)/.4) = 39

Posted: **Sun Jun 14, 2009 4:02 pm**

(4!-4+sqrt(4))/.4 = 55
4!+4!+4+4 = 56
(4!-sqrt(4))/.4 + sqrt(4) = 57
(4!+4)*sqrt(4)+sqrt(4) = 58
(4!/.4)-(4/4) = 59
(4!+4)*sqrt(4)+4 = 60
(4!/.4)+(4/4) = 61
(4!/.4)+4-sqrt(4) = 62
(4!+sqrt(4))/sqrt(4)-sqrt(4) = 63
(4+4)*(4+4) = 64

Posted: **Sun Jun 14, 2009 4:30 pm**

My apologies Daniel

Posted: **Sun Jun 14, 2009 4:54 pm**

Can't see 65
(4!+!4)x(4/sqr4)=66

Posted: **Sun Jun 14, 2009 5:08 pm**

!4 x (!4-sqrt(4)) + sqrt(4) = 65

Posted: **Sun Jun 14, 2009 6:49 pm**

Unless I've missed it, I'm still waiting for 31 to be done properly?

Posted: **Sun Jun 14, 2009 7:17 pm**

Matt Morrison wrote:Unless I've missed it, I'm still waiting for 31 to be done properly?

Are you regarding recurring decimals as invalid? If not, see post at 9:43 am today (assuming that you're in the same time zone as I am).

Posted: **Sun Jun 14, 2009 7:22 pm**

Howard Somerset wrote:
Matt Morrison wrote:Unless I've missed it, I'm still waiting for 31 to be done properly?
Are you regarding recurring decimals as invalid? If not, see post at 9:43 am today (assuming that you're in the same time zone as I am).

Without doubt. Admittedly using recurring decimals are not as utterly horrible as concatenation (which is totally un-mathematic), but still - Kai said it can be done with out any sort of decimals, so I figure that's the rules we ought to be sticking to. Not just because Kai got this quiz started but because you lot are an intelligent bunch and should appreciate the challenge.

Posted: **Sun Jun 14, 2009 7:34 pm**

4x4x4 + sqrt(!4) = 67

Posted: **Sun Jun 14, 2009 7:39 pm**

Matt Morrison wrote:
Howard Somerset wrote:
Matt Morrison wrote:Unless I've missed it, I'm still waiting for 31 to be done properly?
Are you regarding recurring decimals as invalid? If not, see post at 9:43 am today (assuming that you're in the same time zone as I am).
Without doubt. Admittedly using recurring decimals are not as utterly horrible as concatenation (which is totally un-mathematic), but still - Kai said it can be done with out any sort of decimals, so I figure that's the rules we ought to be sticking to. Not just because Kai got this quiz started but because you lot are an intelligent bunch and should appreciate the challenge.

OK then.

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

Posted: **Sun Jun 14, 2009 7:41 pm**

I'll go for an easy one now

4 x 4 x 4 + 4 = 68

Followed by:

(4! - 4/4) x sqrt(!4) = 69

(4! + !4) x 4/sqrt(4) = 70

4! x sqrt(!4) - 4/4 = 71

4! x (4 - 4/4) = 72

Posted: **Sun Jun 14, 2009 7:58 pm**

4 x 4 x 4 + !4 = 73
4! x 4 - 4! + sqrt(4) = 74

Posted: **Sun Jun 14, 2009 8:01 pm**

Matt Morrison wrote:Without doubt. Admittedly using recurring decimals are not as utterly horrible as concatenation (which is totally un-mathematic), but still - Kai said it can be done with out any sort of decimals, so I figure that's the rules we ought to be sticking to. Not just because Kai got this quiz started but because you lot are an intelligent bunch and should appreciate the challenge.

If we're going without concatenation or any sort of decimal, recurring or not, then we've got a lot of holes, namely:
33, 35, 40 to 46, 55, 57, 59, 61, 62 and not just 31 (which has now been done)

Posted: **Sun Jun 14, 2009 8:05 pm**

4! + !4 + 4 - 4 = 33
4! + !4 + 4/sqrt(4) = 35
4! + !4 + 4 + sqrt(!4) = 40
4! + !4 + 4 x sqrt(4) = 41
4! + !4 x 4 / sqrt(4) = 42
4! + 4 x 4 + sqrt(!4) = 43
4! + 4! - sqrt(4) - sqrt(4) = 44
4! + 4! - !4 / sqrt(!4) = 45
4! + 4! - 4 / sqrt(4) = 46
4! + 4! + !4 - sqrt(4) = 55
4! + 4! + sqrt(!4) x sqrt(!4) = 57
4! + 4! + !4 + sqrt(4) = 59
4! + 4! + !4 + 4 = 61
4 x 4 x 4 - sqrt(4) = 62

That should've plugged all the gaps. So it's now 75 and upwards.

Posted: **Sun Jun 14, 2009 8:34 pm**

!4x!4-sqrt!4-sqrt!4=75
!4x!4-sqrt!4-sqrt4=76
!4x!4-sqrt4-sqrt4=77
I'm not sure these are valid under the rule on concatenation.
Can somebody clarify this?

Posted: **Sun Jun 14, 2009 9:04 pm**

Nice one on finding out about !4, I must've missed that. Anyway, once you're done, try not to use it...

Posted: **Sun Jun 14, 2009 9:13 pm**

Liam Tiernan wrote:!4x!4-sqrt!4-sqrt!4=75
!4x!4-sqrt!4-sqrt4=76
!4x!4-sqrt4-sqrt4=77
I'm not sure these are valid under the rule on concatenation.
Can somebody clarify this?

No problem with any of these.
Concatenation mean running two 4s together to make 44.

Posted: **Sun Jun 14, 2009 10:24 pm**

Howard Somerset wrote:
Liam Tiernan wrote:!4x!4-sqrt!4-sqrt!4=75
!4x!4-sqrt!4-sqrt4=76
!4x!4-sqrt4-sqrt4=77
I'm not sure these are valid under the rule on concatenation.
Can somebody clarify this?
No problem with any of these.
Concatenation mean running two 4s together to make 44.

Thanks Howard, thought it might have meant use of brackets.

Posted: **Sun Jun 14, 2009 10:35 pm**

!4 x sqrt(!4) x sqrt(!4) - sqrt(!4) = 78

Posted: **Sun Jun 14, 2009 10:48 pm**

4! x sqrt(!4) + 4 + sqrt(!4) = 79

Posted: **Sun Jun 14, 2009 11:14 pm**

(4!*4)-(4*4) = 80

Posted: **Sun Jun 14, 2009 11:19 pm**

sqrt(!4) x sqrt(!4) x sqrt(!4) x sqrt(!4) = 81

Posted: **Mon Jun 15, 2009 9:21 am**

We have agreed not to go with decimals or concatenation because Kai said we should be able to do it without. Now that Kai has said the same thing about subfactorial, we really should outlaw that too. In which case a whole lot of holes open up. I believe that at least one of decimal/concatenation/subfactorial has been used in each of the following:

26, 31, 35, 38, 39, 40, 41, 42, 45, 51, 53, 55, 59, 61, 65, 66, 67, 70, 71, 73, 75, 76, 77, 78, 79, 81.

So we're still looking for all of these, together with 82 up.

It's certainly more of a challenge without subfactorial, as, having got used to it, I managed to make each of the numbers
1, 2, 3, 4, 6, 9, 24 out of only one 4. And making numbers from 1-100 out of four from that set was proving to be a rather trivial exercise. Now we're without subfactorial, we have to knock 1, 3, 6 and 9 from that list of numbers derivable from just one 4.

Filling in some of those gaps:

4! + sqrt(4) x 4 / 4 = 26
4! + 4 x 4 - sqrt(4) = 38
4! + 4 x sqrt(4) x sqrt(40) = 40
4! + 4 x 4 + sqrt(4) = 42
4! + 4! + 4! - sqrt(4) = 70
4! + 4! + 4! + 4 = 76

leaving 31, 35, 39, 41, 45, 51, 53, 55, 59, 61, 65, 66, 67, 71, 73, 75, 77, 78, 79, and 81 up.

Posted: **Mon Jun 15, 2009 11:11 am**

Howard Somerset wrote:We have agreed not to go with decimals or concatenation because Kai said we should be able to do it without. Now that Kai has said the same thing about subfactorial, we really should outlaw that too. In which case a whole lot of holes open up. I believe that at least one of decimal/concatenation/subfactorial has been used in each of the following:

26, 31, 35, 38, 39, 40, 41, 42, 45, 51, 53, 55, 59, 61, 65, 66, 67, 70, 71, 73, 75, 76, 77, 78, 79, 81.

Awesome, I've got Howard on my side now!

26 had never been a problem for me: 4!+((4+4)/4) i.e. 24 + (8/4) but yeah, still stuck on 31.

I'm sure that once upon a time I knew about subfactorial, but I don't any more. Generally have spent 8 years trying to forget maths, unaware I was going to try and answer this quiz in 2009.

Posted: **Mon Jun 15, 2009 11:14 am**

Hmm, did you just edit your post Howard? When I replied there was no solutions, otherwise I wouldn't have given my 26 obviously. Most odd. That or I can't read.

Posted: **Mon Jun 15, 2009 11:26 am**

Matt Morrison wrote:Hmm, did you just edit your post Howard? When I replied there was no solutions, otherwise I wouldn't have given my 26 obviously. Most odd. That or I can't read.

Coincidence, Matt. After a gap of 110 minutes, my edit was almost simultaneous with your post.

Posted: **Mon Jun 15, 2009 11:28 am**

31= 4!+(4!+4)/4
35= 4!+(4!-sqrt4)/sqrt4
37= 4!+(4!+sqrt4)/sqrt4
38= 4!+(4!+4)/sqrt4

Posted: **Mon Jun 15, 2009 11:36 am**

40= 4!+4!-4-4
66= 4*4*4+sqrt4
96= 4*4*(4+sqrt4)
98= 4!*4+4-sqrt4
97= 4!*4+4/4
95= 4!*4-4/4
76= 4!*4-4!+4

Posted: **Mon Jun 15, 2009 11:36 am**

Neil Zussman wrote:31= 4!+(4!+4)/4

Posted: **Mon Jun 15, 2009 12:22 pm**

(4! - sqrt(4)) x 4 + 4 = 92
(4! - sqrt(4)) x 4 + sqrt(4) = 90
(4! - sqrt(4)) x 4 - sqrt(4) = 86
(4! - sqrt(4)) x 4 - 4 = 84
(4! - 4) x 4 - sqrt(4) = 78
(4! - 4) x 4 + sqrt(4) = 82
(4! - sqrt(4)) x sqrt(4) x sqrt(4) = 88
4! x 4 + sqrt(4) + sqrt(4) = 100
4! x 4 - 4 + sqrt(4) = 94

Leaving 39, 41, 45, 51, 53, 55, 59, 61, 65, 67, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 99

Posted: **Mon Jun 15, 2009 12:48 pm**

Matt Morrison wrote:
Neil Zussman wrote:31= 4!+(4!+4)/4

Thanks Matt.

Just thought I'd share a nice way of making some easy numbers: 10= sqrt(4!*4+4) and 20= sqrt[(4!*4+4)*4]
The first way only uses three fours, so other numbers could be made by adding 4, sqrt4 and 4!, but unfortunately all those numbers have been spotted by simpler methods.
Only the difficult odd numbers remain now, since Howard has been cheeky and done all the easy evens.

It would probably help if we could make numbers such as 3 and 5 from only two 4's. (Or if 1 could make 3 and 5 from 2 4's)

Posted: **Mon Jun 15, 2009 12:56 pm**

I can make a very good approximation to 3, by doing 3= 4-sqrt(sqrt...(sqrt4)))...) which tends to 3, and only uses two 4's.
Similarly 5= 4+sqrt(sqrt...(sqrt4)))...)
We can then do things like
51= 4!+4!+3
45= 4!+4!-3
93= 4!*4-3
99= 4!*4+3
91= 4!*4-5
53= 4!+4!+5
and so on.
But presumably I'll be told I've cheated.
Obviously if anyone does make 3 or 5 using only two 4's, then simply substitute it into my equations above.
(If you can make 7, then 55= 4!+4!+7, and other numbers can be made this way as well.)

Posted: **Mon Jun 15, 2009 1:18 pm**

So we now have left:
39, 41, 45, 51, 53, 55, 59, 61, 65, 67, 71, 73, 75, 77, 78, 79, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 99, 100
Figure I may as well do the top of the house:
100 = (4!*4)+sqrt(4)+sqrt(4)

And a few more (the rest of the even ones):
94 = 4!*4 - 4 + sqrt(4)
92 = 4!*4 - sqrt(4) - sqrt(4)
90 = 4!*4 - 4 - sqrt(4)
88 = 4!*4 - 4 - 4
86 = (4! - sqrt(4))*4 - sqrt(4)
84 = (4! - 4)*4 + 4
82 = (4! - 4)*4 + sqrt(4)
78 = (4! - 4)*4 - sqrt(4)

Leaving:
39, 41, 45, 51, 53, 55, 59, 61, 65, 67, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 99

Posted: **Mon Jun 15, 2009 1:19 pm**

ahaha it looks like I just copied Howard's post!

I was actually making nachos.

Still, still the odd ones to go...

Posted: **Mon Jun 15, 2009 3:20 pm**

I've not been following this too closely, but can someone refer me to where you've done 33 without using decimals.

Posted: **Mon Jun 15, 2009 11:26 pm**

David Williams wrote:I've not been following this too closely, but can someone refer me to where you've done 33 without using decimals.

Well spotted. I've just looked back and can see two attempts at 33, one of which uses decimals and the other uses the subfactorial. So that's another on the to do list.

Posted: **Tue Jun 16, 2009 6:10 pm**

I'm prepared to wager that 33, and doubtless a few others, can't be done under the restrictions you've set yourselves.

Posted: **Tue Jun 16, 2009 6:12 pm**

David Williams wrote:I'm prepared to wager that 33, and doubtless a few others, can't be done under the restrictions you've set yourselves.

I'm still mystified as to what those restrictions actually are, but you're probably right.

Posted: **Tue Jun 16, 2009 8:11 pm**

How much?

Posted: **Tue Jun 16, 2009 8:26 pm**

Kai Laddiman wrote:How much?

Any chance you could clarify the restrictions? It's probably best to specify the minimum (however you want to define that) allowable for it to remain possible.

Posted: **Tue Jun 16, 2009 8:37 pm**

OK, well I completed the puzzle without decimals, powers, integer brackets, trig, gamma function (?!), concatenecation and subfactorials.

Apart from the things I just said (up a little bit), you can use symbols/letters for things like factorial functions as well.

Posted: **Tue Jun 16, 2009 8:50 pm**

I should know this, but isn't there a function (for lack of the proper name I'll call it phi) that adds up all the numbers from 1 to x? i.e. Phi(x)= 1+...+x
Is this allowed? Then 3= Phi(sqrt4) which only uses one 4 to make 3, and makes our task a lot easier.
Also, is my way of making 3 from two 4's by using infinitely many square roots (posted earlier) legal?

Posted: **Wed Jun 17, 2009 9:16 am**

If this is going to go anywhere I think Kai has to say what functions are allowed, rather than what is not allowed. He's clearly using something no-one else is!

Posted: **Wed Jun 17, 2009 12:12 pm**

There are still one or two you can do with normal ops it seems:

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

Posted: **Wed Jun 17, 2009 3:17 pm**

David Williams wrote:If this is going to go anywhere I think Kai has to say what functions are allowed, rather than what is not allowed. He's clearly using something no-one else is!

Well, I did mention it above, but...

(PS scroll down)

Posted: **Wed Jun 17, 2009 3:52 pm**

As in

sf(4) - 4! = 264

264 / 4 / sqrt4 = 33 ?

Personally I've never heard of superfactorials, whereas 44 and .4 are quite familiar.

c4countdown

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