# 1864 Problem



## Tyson (Dec 2, 2011)

A few years ago I wrote here asking for help on a math problem. I'm asking for help again. In the end, I remember Gottlieb saved the day. Anyway...

The goal of the game is to make the numbers 1 to 100 using the digits, 1, 8, 6, 4, in that order. You can do the following things.

Four basic operations and parenthesis:
1 + 8*6 + 4
(1 + 8) / 6 + 4

You can group together numbers:

18 + 64
186 / 4

You can use decimals and factorials:

.1 * .8 * 6! + .4

.18 * 6 + 4!

You can use square roots because square roots don't take digits to write:

sqrt(1 + 8) + 6 + 4

But you can't raise something to the second power, or cube something UNLESS you have the digit. Similarly, you can't take the cube root of something without the digit.

Finally, you can use vinculum. So .111 + .888 + 6 + 4 = 11.

I'm missing four numbers at this point:

41, 68, 86, and 99. Any help would be appreciated.


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## RNewms27 (Dec 2, 2011)

.11 + (sqroot8 *6*4) is 67.99
Sqroot18 + 6 * 4= 40.97


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## Stefan (Dec 2, 2011)

Tyson said:


> *A few years ago* I wrote here asking for help on a math problem.



Not even one


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## vcuber13 (Dec 2, 2011)

its not


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## Stefan (Dec 2, 2011)

RNewms27 said:


> 1* sqroot8 *6*4 close?


 
Close doesn't count.



tozies24 said:


> (1.1111 + 8.8888)*(6+4) = 99 (I hope this is allowed)


 
It's probably not allowed (you can write 0.1 (the line should be above, not below) but 1.1 would use 1 twice) and it's also wrong (you get 100 instead of 99).


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## Stefan (Dec 2, 2011)

Hm, hang on...

\( 99 = \sqrt{\sqrt{.\overline{1}^{-8}}} - 6 + 4! \)

</shamelessripoff>


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## Robert-Y (Dec 2, 2011)

\( 99 = -\sqrt{\sqrt{.\overline{1}^{-8}}} + 6!/4 \)

Just another solution for 99


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## cubernya (Dec 2, 2011)

I'm searching for 41...this is hard


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## Robert-Y (Dec 2, 2011)

\( 68 = {6!/{\sqrt{\sqrt{\sqrt{.1^{-8}}}}}}-4 \)

Kinda close :S


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## Sa967St (Dec 2, 2011)

Robert-Y said:


> \( 68 = {6!/{\sqrt{\sqrt{\sqrt{.1^{-8}}}}}}-4 \)
> 
> Kinda close :S


86= 1*(6!/8)-4 
Kinda close


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## Stefan (Dec 2, 2011)

1, 8, 6, 4, *in that order*


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## cubernya (Dec 2, 2011)

Sa967St said:


> 86= 1*(6!/8)-4
> Kinda close


 
1684? Don't think so

EDIT: Ninja'd


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## Sa967St (Dec 2, 2011)

Stefan said:


> 1, 8, 6, 4, *in that order*





theZcuber said:


> 1684? Don't think so
> 
> EDIT: Ninja'd


I know, that's why I said it was "kind of close".


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## Robert-Y (Dec 2, 2011)

\( 86 = {\sqrt{\sqrt{(1*86)^4}}} \)

Lol why didn't we think of this before...


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## cubernya (Dec 2, 2011)

Robert-Y said:


> \( 86 = {\sqrt{\sqrt{(1*86)^4}}} \)
> 
> Lol why didn't we think of this before...


 
Oh wow. I feel stupid.


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## cuBerBruce (Dec 2, 2011)

Just fix the order. 86 = (1/8)*6!-4. Or 86 = sqrt(sqrt((1*86)^4)). (Ninja'd on the latter one.)


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## Robert-Y (Dec 2, 2011)

This probably doesn't count and I don't actually know how to write it out properly but here it goes:

41 = ((sqrt(sqrt(0.111...^-8)))+(sqrt^infinite(6)))/(sqrt(4))


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## zmikecuber (Dec 2, 2011)

29=1+8+6+4


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## tozies24 (Dec 2, 2011)

Robert-Y said:


> This probably doesn't count and I don't actually know how to write it out properly but here it goes:
> 
> 41 = ((sqrt(sqrt(0.111...^-8)))+(sqrt^infinite(6)))/(sqrt(4))


 
Well you cant really take an infinite root of something without a limit.. and I dont think thats allowed here.

EDIT: Also, the way you have it written is just not right, since you would want something like the nth root where n -> infinity.


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## Sa967St (Dec 2, 2011)

Robert-Y said:


> \( 86 = {\sqrt{\sqrt{(1*86)^4}}} \)
> 
> Lol why didn't we think of this before...


 or 86 = \( {\sqrt{(1*86)^{\sqrt{4}}}} \)



cuBerBruce said:


> Just fix the order. 86 = (1/8)*6!-4.


Ah yes, I realized that too afterwards.


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## tozies24 (Dec 2, 2011)

how did you get 31? I am just wondering since 31 and 41 are both prime and somewhat similiar in size.


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## Stefan (Dec 2, 2011)

tozies24 said:


> how did you get 31? I am just wondering since 31 and 41 are both prime and somewhat similiar in size.



31 is easy: 
-1 + 8 + 6*4


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## qqwref (Dec 2, 2011)

41 = 1 + 8 * mean(6,4) 

I know, not a very nice solution, but it's all I have so far.


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## Tyson (Dec 2, 2011)

41 = sqrt( 1 + ( 8! / ( 6 * 4 ) ) )

Courtesy of someone at Zynga. Last on is 68.


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## tozies24 (Dec 2, 2011)

(-1 + 8) * 6 - nth root of 4 = 41 (if we are able to use limits...)

EDIT: ninja i guess. thats a pretty good solution


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## qqwref (Dec 2, 2011)

This is ugly, but:

68 = floor(sqrt(18)) + 64


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## Carrot (Dec 2, 2011)

Didn't we have the same problem some years ago? or was it 1623?


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## Tim Major (Dec 2, 2011)

Is it rounded? does 10.9 = 11, or must it be accurate?


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## Carrot (Dec 2, 2011)

Tim Major said:


> Is it rounded? does 10.9 = 11, or must it be accurate?


 
it must be accurate, last time I had something that was 0.02ish away from the correct number, that didn't count


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## Sebastien (Dec 2, 2011)

well, 0.1 * 8^0 * 6! -4 is not allowed, right?


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## Carrot (Dec 2, 2011)

Sébastien_Auroux said:


> well, 0.1 * 8^0 * 6! -4 is not allowed, right?


 
no, you inserted a zero, but you could use the infity root hack?


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## cuBerBruce (Dec 2, 2011)

#include <math.h>
int main () { return sqrt(sqrt((double)(1 << 8)))+64; }


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## Tyson (Dec 2, 2011)

Sorry, I got this one last night.

68 = sqrt( .1111 * 8! + 6 * 4!)


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## Stefan (Dec 2, 2011)

Dammit I did think about sqrt-ing at the very end but apparently gave up way too easily.

Btw, why are you doing this? I mean, it's a neat riddle for a little while, but I wouldn't have the patience to go for all 100 and I certainly wouldn't do it year after year. Are you providing sample solutions for that school, going after some prize, or something else? Just curious.


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## Litz (Dec 2, 2011)

I hope it's just for fun since there's better ways to do this than by hand. It's a cool challenge but I honestly wouldn't have the patience to do it. If I was forced to do it I'd probably just code a solver.


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## Sa967St (Dec 2, 2011)

Just wondering, what's the significance of 1864? I remember last year when it was 1623 you said that it was Pascal's year of birth.


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## cubersmith (Dec 2, 2011)

vcuber13 said:


> its not


 
Explain


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## Stefan (Dec 2, 2011)

cubersmith said:


> Explain


 
He replied to tozies24 who then deleted his post (you can still see it quoted by me if you want).


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## tozies24 (Dec 2, 2011)

Stefan said:


> He replied to tozies24 who then deleted his post (you can still see it quoted by me if you want).


 
I thought that I would be quick and delete mine but it wasn't quick enough. I realized that mine was completely wrong.

On topic: You guys mentioned significance to the numbers chosen. Today I wrote down 1492 and just started to go through some of the possibilites up to 40. 
Is there any restriction that is put on the numbers that you choose? I'm assuming that as long as you don't have a 0 and as long as you don't have more than one 1 that it is possible to get every number from 1 to 100.


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## qqwref (Dec 2, 2011)

Sébastien_Auroux said:


> well, 0.1 * 8^0 * 6! -4 is not allowed, right


Actually.....
68 = (sqrt(sqrt(sqrt(.1^8))) * 6!) - 4

PS: Nice solution, Tyson!


EDIT: lol, I just came up with the Zynga guy's 41 solution again.


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## Carrot (Dec 2, 2011)

Sa967St said:


> Just wondering, what's the significance of 1864? I remember last year when it was 1623 you said that it was Pascal's year of birth.


 
I remember solving the 1623 problem :3 I think I was missing 1 or 2 numbers in the end


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## aaronb (Dec 2, 2011)

Sa967St said:


> Just wondering, what's the significance of 1864? I remember last year when it was 1623 you said that it was Pascal's year of birth.


 
Well I'm assuming Tyson is from America since in his location, it says California; and I generally associate 1864 with the U.S. Civil War; so maybe it is something related to that?


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## vcuber13 (Dec 2, 2011)

George Boole died Dec 8


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## Tim Major (Dec 3, 2011)

Stefan said:


> 31 is easy:
> -1 + 8 + 6*4


 
-.111 + 8! / 6^4 = 31 
(I got yours first)
I'm doing it in a notebook right now (I'm in the middle of no where with nothing to do...).
I got something similar to Rob's 86 solution, but I didn't get 86 until I saw his post >_<

47, 48 and 49 using;
[sqrt [sqrt (8*6)^4]]
and combinations of -1, +1, *1.

I'm just doing random combinations, to get a random number, then I change the 1 at the start.
I have 1-33, and 58 total, though I'm not currently stuck on any in particular, and some I've left untouched.
Congrats for finishing it :tu


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