Four 4s

[Howard Somerset]

Next one..........not so easy.

No problem if we use recurring decimals.

Code: Select all

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

[Liam Tiernan]

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

[Howard Somerset]

44-(4/.4)=37

No. This makes 34.

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

[Liam Tiernan]

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.

[Daniel O'Dowd]

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

[Jason Larsen]

16 = 4 + 4 + 4 + 4.

[Innis Carson]

(!4)

What function is this? Looks pretty helpful.

[Liam Tiernan]

(!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

[Peter Mabey]

(!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.

[Daniel O'Dowd]

(!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

[Innis Carson]

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

[James Hall]

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.

[Howard Somerset]

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

[James Hall]

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

[James Hall]

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

[James Hall]

(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

[Liam Tiernan]

My apologies Daniel

[Paul Erdunast]

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

[Howard Somerset]

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

[Matt Morrison]

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

[Howard Somerset]

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).

[Matt Morrison]

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.

[Ray Folwell]

4x4x4 + sqrt(!4) = 67

[Howard Somerset]

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

[Howard Somerset]

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

[Ray Folwell]

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

[Howard Somerset]

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)

[Howard Somerset]

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.

[Liam Tiernan]

!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?

[Kai Laddiman]

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

[Howard Somerset]

!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.

[Liam Tiernan]

!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.

[Innis Carson]

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

[Howard Somerset]

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

[Steve Durney]

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

[Howard Somerset]

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

[Howard Somerset]

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.

[Matt Morrison]

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.

[Matt Morrison]

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.

[Howard Somerset]

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.

[Neil Zussman]

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

[Neil Zussman]

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

[Matt Morrison]

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

[Howard Somerset]

(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

[Neil Zussman]

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)

[Neil Zussman]

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.)

[James Hall]

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

[James Hall]

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

I was actually making nachos.

Still, still the odd ones to go...

[David Williams]

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

[Howard Somerset]

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.

[David Williams]

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

[Charlie Reams]

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.

[Kai Laddiman]

How much?

[Michael Wallace]

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.

[Kai Laddiman]

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.

[Neil Zussman]

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?

[David Williams]

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!

[Simon Myers]

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

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

[Kai Laddiman]

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)

[David Williams]

As in

sf(4) - 4! = 264

264 / 4 / sqrt4 = 33 ?

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