# 1623 problem



## Tyson (Dec 20, 2010)

Hi Everyone,

So when I was in middle school, every December, we would get the "New Year problem," or, back in those days, it was the 1996 problem or the 1997 problem. The goal was to use the digits 1, 9, 9, 6 in that order to make the numbers 1 through 100. There was a lot more flexibility than normal... so we were able to use +, -, *, /, and other things like sqrt, factorial, decimals, etc.

Naturally, when 2000 rolled around, this problem was not really doable. So they've been picking different years. This year, the digits are 1623, the year Pascal was born. The digits simply have to be put in the correct order... so some examples:

1 = 1^623 (you can combine digits too)
2 = 1 + 6 - 2 - 3
21 = 16 + 2 + 3
25 = 1 + ((6 * 2) / 3)!
80 = .1 * 6! + 2^3

I'm having trouble on a few numbers. Let me know if you can figure out solutions to the following:

41, 44, 83, 87, 92, 93, 94, and 98.

I'd imagine there will be some people who get the rules wrong for a bit. Basically, the numbers have to be written in that order. You can use things like decimals, square roots, because they don't require extra digits. But a cube root would require an extra digit, so you can't use that, though, you could theoretically use a 6th root with the 6...

There are some creative tricks that I've used, but I won't spell those out in case someone actually wants to try the problem. Just as a note, this was much easier in the 90's because 9's are so useful! You have 9, sqrt 9 = 3, factor of sqrt 9 = 6... so delicious!

But they're all low digits this year, so I'm having a bit of trouble. Let me know if you find any, or try the entire problem yourself!


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## Sa967St (Dec 20, 2010)

Ooh I used to do these in middle school too. I'll give this one a try.

43/100 so far


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## einstein00 (Dec 20, 2010)

87 = ceil(-floor(tan 1.6) * 2 * arcsec 3)

all the others:

while(current != goal) current = floor(162.3*rand);


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## Feinster (Dec 20, 2010)

Our class did the _exact_ same thing when we were 11 or 12! We used to love 9s for the exact reason that you posted, they're so easy to manipulate! Anyway, I'll try to find some, I'll post them here if I find any.


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## mr. giggums (Dec 20, 2010)

87=261/3


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## Feinster (Dec 20, 2010)

mr. giggums said:


> 87=261/3



You have to use them in their original order.


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## Tyson (Dec 20, 2010)

einstein00 said:


> 87 = ceil(-floor(tan 1.6) * 2 * arcsec 3)
> 
> all the others:
> 
> while(current != goal) current = floor(162.3*rand);


 
Middle school math... so back when I did this in school, we weren't using trig functions or ceiling and floor. Otherwise, it's pretty easy.


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## Tyson (Dec 20, 2010)

Sa967St said:


> Ooh I used to do these in middle school too. I'll give this one a try.
> 
> 43/100 so far
> 
> ...


 
Haha, you're cute. 4 * 3 * 2 * 1...

By the way, I have found a solution for 50.


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## Sa967St (Dec 20, 2010)

Tyson said:


> Haha, you're cute. 4 * 3 * 2 * 1...


my bad ._.
I accidentally punched in 4x3x1 instead of 4x3x2 in my calculator.


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## shelley (Dec 20, 2010)

You needed a calculator?


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## Tyson (Dec 20, 2010)

It's better than some of Bob's students... "What's negative one plus five?!? I NEED A CALCULATOR!"

I found a solution to 93.

So all I need now are the following:

41, 44, 83, 87, 92, 94, and 98.

*UPDATE* I found a solution for 83.


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## Sa967St (Dec 20, 2010)

41 = -1+(6^2)+(3!)


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## Tyson (Dec 20, 2010)

Sa967St said:


> 41 = -1+(6^2)+(3!)


 
NICE! I'm surprised I missed that one actually. I'm trying a lot of the "hard core" stuff right now.


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## cuBerBruce (Dec 20, 2010)

98 = floor((ln(16!)+2)*3)


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## Tyson (Dec 20, 2010)

cuBerBruce said:


> 98 = floor((ln(16!)+2)*3)


 
If you use floor, ceiling, then it's all really really trivial. I'm going to say we should try to do this without functions that require letters, so no trig functions either, and no log.

Still have 44, 87, 92, 94, and 98.

Spoilers... if anyone wants to see what I've done, or check my work:

https://spreadsheets.google.com/ccc?key=0AsCSyGmELaOedElpQllQUFZlVkw0NlBObWtpeEVQMHc&hl=en


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## cmhardw (Dec 20, 2010)

Tyson said:


> Still have 44, 87, 92, 94, and 98.


 
44 = 1 + 6!! - 2 - 3
92 = -1 + 6!! * 2 - 3
94 = 1 + 6!! * 2 - 3
98 = -1 + 6!! * 2 + 3

Where !! is the double factorial.


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## Sa967St (Dec 20, 2010)

92= (√16)x23


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## Tyson (Dec 20, 2010)

Sa967St said:


> 92= (√16)x23


 
Wow, go Sarah! sqrt(16) * 23 is so elegant.

I've never heard of this double factorial function. I'll have to re-write a lot of the (3!)! = 720 type stuff.


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## Stefan (Dec 20, 2010)

Tyson said:


> I'll have to re-write a lot of the (3!)! = 720 type stuff.


 
Isn't that something quite different?


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## Tyson (Dec 20, 2010)

Stefan said:


> Isn't that something quite different?


 
I'm not sure I understand you. We're saying 3!! = 3 by double factorial function, but (3!)! = 720, because 3! = 6, and 6! = 720.

Got 87 using double factorial function... if anyone can improve on the highlighted solutions, without using a double factorial, it would be greatly appreciated.


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## Sa967St (Dec 20, 2010)

87= (-1+(6/.2))x3


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## Diniz (Dec 20, 2010)

Sa967St said:


> 87= (-1+(6/.2))x3


 
87 = 261/3


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## Stefan (Dec 20, 2010)

Tyson said:


> I'm not sure I understand you. We're saying 3!! = 3 by double factorial function, but (3!)! = 720, because 3! = 6, and 6! = 720.


 
Ok, sorry. It looked like you wanted to just re-write (x!)! as x!! to make it look cleaner and I thought you had overlooked the difference. It did feel odd, because it wouldn't save much even if it worked and you wouldn't have written the parentheses in the first place, but I didn't see what else you could've meant. How did you mean it then? Can you give an example?


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## Sa967St (Dec 20, 2010)

I'm stuck after 81/100.
My answers:


Spoiler



1 = 1^623
2 = 1+6-2-3
3 = 1+((6x2)/3!)
4 = (1^62)+3
5 = 1+((6x2)/3)
6 = 1+6-2+3
7 = (1^62)+3!
8 = 16-2-(3!)
9 = 1((sqrt(6^2))+3)
10 = -1+6+2+3
11 = 16-2-3
12 = 1+6+2+3
13 = 1+6+(2x3)
14 = (1x6)+(2^3)
15 = 1+6+(2^3)
16 = -1-6+23
17 = (-1x6)+23
18 = 1-6+23
19 = (1x6)+23
20 = 16-2+3!
21 = 16+2+3
22 = ((sqrt16)^2)+(3!)
23 = -1+(6(-2+(3!)))
24 = 16+2+3!
25 = 1+(6(-2+(3!)))
26 = -1+(((6-2)!)+3)
27 = 1(((6-2)!)+3)
28 = -1+6+23
29 = 16x2-3
30 = 1+6+23
31 = 1+((6^2)-(3!))
32 = -1+(6^2)-3
33 = 1(6^2-3)
34 = 1+(6^2)-3
35 = (16x2)+3
36 = (sqrt16)(2x(3!))
37 = 1+(((6/2)!)(3!))
38 = -1+(6^2)+3
39 = 16+23
40 = 1+(6^2)+3
41 = -1+(6^2)+)3!)
42 = 1((6^2)+(3!))
43 = 1+(6^2)+3!
44
45
46 = (1+6)^2 -3
47 = -1+(6(2^3))
48 = (sqrt16)x2x(3!)
49 = .1(6!)-23
50
51
52 = (1+6)^2+3
53
54 = 162/3
55 = -1+62-(3!)
56 = 1x(62-(3!))
57 = 1+62-(3!)
58 = -1+62-3
59 = 1(62-3)
60 = 1+62-3
61
62
63 = -1+(6-2)^3
64 = 16x(-2+(3!))
65 = 1+(6-2)^3
66 = .1(6!)-(2x3)
67 = .1(6!)-2-3
68 = .1(6!)+2-(3!)
69 = 1+62+3!
70 = 1(6+(2^(3!)))
71 = .1(6!)+2-3
72 = .1(6!)(-2+3)
73 = .1(6!)-2+3
74
75 = (1+(6-2)!)x3
76 = .1(6!)-2+(3!)
77 = .1(6!)+2+3
78 = .1(6!)+(2x3)
79
80 = .1(6!)+(2^3)
81
82
83 = (16/.2)+3
84 = .1(6!)+(2)(3!)
85
86
87 = (-1+(6/.2))x3
88
89 = -1+((6/.2)x3)
90 = 1((6/.2)x3)
91 = 1+((6/.2)x3)
92 = (sqrt16)x23
93 = (1+(6/.2))x3
94
95 = .1(6!)+23
96 = ((sqrt16)^2)(3!)
97
98
99
100



edit: I didn't think of using repeating decimals. :/


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## mr. giggums (Dec 20, 2010)

Diniz said:


> 87 = 261/3


 
They have to be in order. I made the same mistake in post 5.


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## Ranzha (Dec 20, 2010)

Diniz said:


> 87 = 261/3



I do believe the numbers have to be in sequential order.

EDIT: And I have been ninja'd my mr. giggums here. Good work, sir. -Shakes hand.-


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## Diniz (Dec 20, 2010)

Ranzha V. Emodrach said:


> I do believe the numbers have to be in sequential order.
> 
> EDIT: And I have been ninja'd my mr. giggums here. Good work, sir. -Shakes hand.-



Ahhh! I imagined that no one would miss an easy one lol


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## Julian (Dec 21, 2010)

Has 100 been done?


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## Ranzha (Dec 21, 2010)

Julian said:


> Has 100 been done?


 
Yes. Check the Google doc. It has all of the solutions.
It'd be better, in my opinion, to omit the use of repeating decimals, such as .111 or stuff.


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## Tyson (Dec 21, 2010)

Sa967St said:


> 87= (-1+(6/.2))x3


 
Gah, nice. I'm surprised I overlooked this one.

In my opinion, and of course, we all grew up with this problem differently, decimals and repeating decimals are fair.

Though, it would be nice to do 44, 94, and 98 without using the double factorial.

https://spreadsheets.google.com/ccc?key=0AsCSyGmELaOedElpQllQUFZlVkw0NlBObWtpeEVQMHc&hl=en

You can now edit sheet 2 and share any solutions you find particularly elegant or creative.


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## cmhardw (Dec 21, 2010)

Tyson said:


> Gah, nice. I'm surprised I overlooked this one.
> 
> In my opinion, and of course, we all grew up with this problem differently, decimals and repeating decimals are fair.
> 
> ...


 
The 68 line doesn't work. What's written is equal to 70.
68 = .1 * 6! - (-2 + 3!) should do it.


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## Carrot (Dec 21, 2010)

cmhardw said:


> The 68 line doesn't work. What's written is equal to 70.
> 68 = .1 * 6! - (-2 + 3!) should do it.


 
I used 68=1*62+3!
much more simple solution


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## Tyson (Dec 21, 2010)

Tord said:


> Gasp, #*44* seems quite difficult. _Must.. not.. adjust.. decimals.._
> 
> ,111... * (6! * (,22 + ,33)) _≈ 44_
> 
> Oh well, better move on to *#94* and *#98*. More fun awaits! :>


 
Not quite 44 though. So that doesn't work... you can't use .22, because the notation for that requires two digits of 2. But .11111.... is just .1 with the bar over it.


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## Sa967St (Dec 21, 2010)

98 is 7x14, so if you could somehow get 14 out of 2 and 3 you could do 98=(1+6)x14.


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## MichaelP. (Dec 21, 2010)

Sa967St said:


> 98 is 7x14, so if you could somehow get 14 out of 2 and 3 you could do 98=(1+6)x14.


 
(1+6)*2(cubed)+3! = 98. Unless cubing something requires I use a 3.


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## Tyson (Dec 21, 2010)

MichaelP. said:


> (1+6)*2(cubed)+3! = 98. Unless cubing something requires I use a 3.


 
Squaring, cubing, raising to any power requires the use of a digit.

In terms of roots, square root is legal because it doesn't require a digit but cube roots are not, because you need a 3.


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## Stefan (Dec 21, 2010)

MichaelP. said:


> (1+6)*2(cubed)+3! = 98. Unless cubing something requires I use a 3.


 
I don't know what's more obvious. That you can't do that, or that it's wrong.


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## Tord (Dec 21, 2010)

Tyson said:


> Not quite 44 though. So that doesn't work... you can't use .22, because the notation for that requires two digits of 2. But .11111.... is just .1 with the bar over it.



Oh, sigh..

I thought my choice of wording combined with the withering gray colour indicated what you just said.
OH, WHY YES: Correct - _,111..._ is indeed ".1 with the bar over it.", neat appliance of acumen. >.>"


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## MichaelP. (Dec 21, 2010)

Stefan said:


> I don't know what's more obvious. That you can't do that, or that it's wrong.


 
Ah, sorry Stefan. (1+6)*(2(cubed)+3!) = 98 Fixed. It's against the rules anyway so I'll try again.


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## Kynit (Dec 21, 2010)

MichaelP. said:


> Ah, sorry Stefan. (1+6)*(2(cubed)+3!) = 98 Fixed. It's against the rules anyway so I'll try again.


 
(1+6)*(2^2+3!)
7*(4+6)
7*10
70

Am I doing something wrong?


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## Tord (Dec 21, 2010)

Kynit said:


> (1+6)*(2^2+3!)
> 7*(4+6)
> 7*10
> 70
> ...



The cube of a number is in the power of three, so it is 2^3.


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## MichaelP. (Dec 21, 2010)

Kynit said:


> (1+6)*(2^3+3!)
> 7*(*8*+6)
> 7*14
> 98
> ...



Fixed


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## cyoubx (Dec 21, 2010)

FINALLY!!!

44 = -(1*6-2) +((3!)!!)


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## cyoubx (Dec 21, 2010)

Tord said:


> Hardly.
> 
> 
> 
> ...


 
1. Oh...didn't see that.
2. Okay

How does my answer not work mathematically?


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## reThinking the Cube (Dec 21, 2010)

94 = -1+.(6!)+23


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## Sa967St (Dec 21, 2010)

reThinking the Cube said:


> 94 = -1+.(6!)+23


.(6!) ≠ 72
I don't think that would follow the rules anyway.



Tord said:


> Gasp, #*44* seems quite difficult. _Must.. not.. adjust.. decimals.._
> 
> ,111... * (6! * (,22 + ,33)) _≈ 44_



1 + ((6!)((.2)(.3)) ≈ 44

getting closer x)


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## Kynit (Dec 21, 2010)

Tord said:


> The cube of a number is in the power of three, so it is 2^3.


 
DURRRR thanks.


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## qqwref (Dec 22, 2010)

Sa967St said:


> 98 is 7x14, so if you could somehow get 14 out of 2 and 3 you could do 98=(1+6)x14.


 
((1+6)^2)*ceil(sqrt(3)) =98 
((1+6)^2)*fib(3) = 98 
Yeah, I know these aren't really legit solutions.


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## maggot (Dec 22, 2010)

given sarahs train of thought, 44 = 11 x 4, so if you can make 11 and 4 out of 1, 6, 3, and 2 then you can find a soln. 

on topic : i remember also doing this in school and was able to come up with soln using functions. using pure computations, i was not able to come up with 44. i tried for about 2 hours. im going to give more effort...((1+6)^2)*fib(3) = 98 i also came up with a fib 3 soln for 98. that should be as legit as sqrt lol. i think its much cleaner than 6!! ! LOL


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## qqwref (Dec 22, 2010)

Actually, considering ceil and floor can be written with just lines (no numbers, no letters), I'd say they are arguably just as clean as (if not cleaner than) double factorials.


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## maggot (Dec 22, 2010)

agreed
lol 43, close to 44, fib 16 /23


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## Rinfiyks (Dec 22, 2010)

Is gamma function allowed?


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## maggot (Dec 22, 2010)

{1 + ( (sqrt (sqrt (sqrt (sqrt 6) )^(-1) )} x 23 ~ 44 (if you can find another way to write reciprocals)


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## Sa967St (Dec 22, 2010)

maggot said:


> given sarahs train of thought, 44 = 11 x 4, so if you can make 11 and 4 out of 1, 6, 3, and 2 then you can find a soln.


sqrt16 gives you a 4, but I have no idea how to get 11 out of a 2 and 3.


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## maggot (Dec 22, 2010)

Rinfiyks said:


> Is gamma function allowed?


do you have a soln with gamma used ? assuming gamma function where = (n-1)! ?


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## reThinking the Cube (Dec 22, 2010)

44= 1*Σ6+23


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## qqwref (Dec 22, 2010)

What function do you mean when you say "Σ6"?


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## uberCuber (Dec 22, 2010)

qqwref said:


> What function do you mean when you say "Σ6"?


 
he was using that for 6+5+4+3+2+1


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## reThinking the Cube (Dec 22, 2010)

uberCuber said:


> he was using that for 6+5+4+3+2+1


 
Yeah, that is right.

Sum of Natural Numbers
The sum of the natural numbers is the value obtained when a selection of the numbers (beginning from 1) are added together. For example the sum of the first 6 natural numbers is 21 because:

1 + 2 + 3 + 4 + 5 + 6 = 21


If the sum of a longer number sequence is required (say the sum of the first 100 numbers) this process can become tedious. There is, luckilly, a formula for summing the first n natural numbers. The symbol Σn (the Greek capital Sigma followed by n) is used to denote the sum of the first n natural numbers. The formula can then be written:

Σn = n × (n + 1) / 2


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## Stefan (Dec 22, 2010)

reThinking the Cube said:


> Sum of Natural Numbers
> The sum of the natural numbers is the value obtained when a selection of the numbers (beginning from 1) are added together. For example the sum of the first 6 natural numbers is 21 because:
> 
> 1 + 2 + 3 + 4 + 5 + 6 = 21
> ...



Where was that copied from?


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## reThinking the Cube (Dec 22, 2010)

Stefan said:


> Where was that copied from?



Yeah, thats right too. I didn't feel like putting much effort into answering a rather obvious question, so I copy/pasted it from an Intro to Numbers page, that I googled. qq's question reminded me of the time Albert E. forgot what "E" stood for.


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## qqwref (Dec 22, 2010)

Well, I was asking because I haven't seen the sigma notation before (not how you are writing it anyway), so if it's a standard notation at all it is rarely used. It is quite possible that whoever wrote the page made up that notation, and only used it to save space in the quoted article. MathWorld seems to think the standard notation for 1+...+n is \( T_n \) - you can argue with them if you want.

EDIT: I took a look at the page you mentioned and it looks pretty amateurish. They've clearly put some work into the page but it doesn't seem like the authors are serious mathematicians. I really doubt this is a normal notation (and of course if it isn't there is no reason anyone would be expected to know it).


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## einstein00 (Dec 22, 2010)

reThinking the Cube said:


> Albert E. forgot what "E" stood for.


 
lol wut?


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## reThinking the Cube (Dec 22, 2010)

qqwref said:


> Well, I was asking because I haven't seen the sigma notation before (not how you are writing it anyway), so if it's a standard notation at all it is rarely used. It is quite possible that whoever wrote the page made up that notation, and only used it to save space in the quoted article. MathWorld seems to think the standard notation for 1+...+n is \( T_n \) - you can argue with them if you want.
> 
> EDIT: I took a look at the page you mentioned and it looks pretty amateurish. They've clearly put some work into the page but it doesn't seem like the authors are serious mathematicians. I really doubt this is a normal notation (and of course if it isn't there is no reason anyone would be expected to know it).



Here is another one of those *amateurish* sources - http://en.wikipedia.org/wiki/Summation


> Informal writing sometimes omits the definition of the index and bounds of summation when these are clear from context,



Anyway - is this your preference?

\( 44 = 1*T_6+23 \)


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## qqwref (Dec 22, 2010)

reThinking the Cube said:


> Here is another one of those *amateurish* sources - http://en.wikipedia.org/wiki/Summation


Take a closer look at what you quoted: Wikipedia suggests the completely different notation of Σi. By their notation, Σ6 would mean "sum over some set of the function "6"", which would be equivalent to 6 times the size of the set. And that isn't what you wanted.



reThinking the Cube said:


> Anyway - is this your preference?
> 
> \( 44 = 1*T_6+23 \)


This is how I'd say your idea should be written, yeah. I think we agreed that we should try to avoid letters in our solution, though, so this solution wouldn't really work.


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## Julian (Dec 22, 2010)

-1 + 6 * 2 gives you 11. Now go make 4 out of 3


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## qqwref (Dec 22, 2010)

ceil(sqrt(sqrt(sqrt(sqrt((3!)!))!)))


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## reThinking the Cube (Dec 22, 2010)

qqwref said:


> Take a closer look at what you quoted: Wikipedia suggests the completely different notation of Σi. By their notation, Σ6 would mean "sum over some set of the function "6"", which would be equivalent to 6 times the size of the set. And that isn't what you wanted.



You are misinterpreting what it means to omit the index and bounds of summation.



qqwref said:


> This is how I'd say your idea should be written, yeah. I think we agreed that we should try to avoid letters in our solution, though, so this solution wouldn't really work.



Avoid letters? I don't think anyone agreed that we should try to avoid math SYMBOLS (such as sigma, sqrt, or !) in our solution. According to Michael Gottlieb and MathWorld, the \( T \) symbol is the standard notation for 1+...+n. You can argue with THEM if you want.


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## qqwref (Dec 22, 2010)

reThinking the Cube said:


> You are misinterpreting what it means to omit the index and bounds of summation.


Take a look at the wikipedia page you linked to: it suggests
\( \sum x_i^2 := \sum_{i=1}^n x_i^2 \)
Following the pattern, the sum for the triangular number of 6 would normally be written \( \sum_{i=1}^6 i \), so in this notation it would be \( \sum i \), not \( \sum 6 \).



reThinking the Cube said:


> Avoid letters? I don't think anyone agreed that we should try to avoid math SYMBOLS (such as sigma, sqrt, or !) in our solution. According to Michael Gottlieb and MathWorld, the \( T \) symbol is the standard notation for 1+...+n. You can argue with THEM if you want.


When the topic of letters was brought up it was meant for stuff like sin, ln, etc. T is definitely a letter (yes, letters are often used for notational purposes in math, but they are still letters). Sigma is on the line; \( \sqrt{} \) and ! are clearly not letters.

Just wondering, but why are you treating this as some kind of battle? I'm only trying to show you that Σ6 isn't proper notation, so that you can better help with the 1623 problem.


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## Carrot (Dec 23, 2010)

Damn this is tough... I'm missing 4 numbers right now (I'm not telling you what 4 numbers it is )

I'll post my results whenever I find one more =)


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## reThinking the Cube (Dec 23, 2010)

qqwref said:


> When the topic of letters was brought up it was meant for stuff like sin, ln, etc. T is definitely a letter (yes, letters are often used for notational purposes in math, *but they are still letters). Sigma is on the line * [sic]; \( \sqrt{} \) and ! are clearly not letters.



Your missing the point. Maybe this will help you.

http://kwarc.info/publications/papers/kw1_notationSurvey.pdf


> This paper summarizes various examples that illustrate the ambiguity of notations.
> In particular, we pointed out alternative notations that are used to present the
> same mathematical concept, i.e. mathematical content expression with the same
> meaning. In addition, we exemplied mathematical expression, which represent
> ...


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## qqwref (Dec 23, 2010)

Maybe I am missing the point - what is it? I took a casual look at the paper but I don't see the relevance. If you are trying to say that alternative notations exist, I agree, but that doesn't mean that a notation created by a random person should automatically be accepted as standard. I think one implicit goal of the 1623 project is to create solutions that everyone with a math background can follow - using rare or very informal notation seems to go against this.

What did you mean by the [sic]?


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## abctoshiro (Dec 23, 2010)

Okay, rethinking the cube (and qqwref), get back on topic please. I'm not mad or angry or <insert negative emotion> but I feel like you are straying away. 

I think that symbols without letters should be used. Anyway,

On topic: This is hard. 98 and 44 stump me the most. I'm still thinking about a solution without a double factorial. I'll post if I find one.


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## qqwref (Dec 23, 2010)

I'd love to get back on topic, but I think reThinking might be trying to make a relevant point, and I haven't figured out what it is yet.

98 and 44 are pretty tricky. Nothing yet (other than what I posted above).


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## keemy (Dec 23, 2010)

\( 44 = -1 + \binom {6}{2}3 \)

idk if you don't want to use binomial coef /n choose k but it's definitely less obscure than double factorial.


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## Tyson (Dec 23, 2010)

Nice Keemy, it is less obscure than the double factorial no doubt. Unfortunately, this problem is given to sixth graders, and sixth graders actually don't learn normal factorial until after the New Year at this school. So really, to do the problem completely kosher, you're limited to the basic four operations +, -, *, /, decimals, repeating decimals, and factorials. Also powers, but you need a digit, and square root, but you don't need a digit. And you could use a cube root or something if you wanted, but you would need to utilize a digit, so perhaps you can use the 6th root. I wonder if that would help...

But yeah, obviously the more simple the solution, the more reasonable it is for a 6th grader to get.


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## Robert-Y (Dec 23, 2010)

I'm not really sure if this is helpful, but I managed to make 440...

-(sqrt((.1)^-6))+(2*((3!)!))


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## Carrot (Dec 23, 2010)

keemy said:


> \( 44 = -1 + \binom {6}{2}3 \)
> 
> idk if you don't want to use binomial coef /n choose k but it's definitely less obscure than double factorial.


 
When you write that out, you can't really tell wether 6 or 2 is first


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## Carrot (Dec 23, 2010)

Tord said:


> It is written 6C2 :>


 
but then you are using a letter


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## trying-to-speedcube... (Dec 23, 2010)

\( 44=.1^{-{\sqrt{\sqrt{\sqrt{\ldots{\sqrt{6}}}}}}}/.2-3! \)

DISCLAIMER Odder did this, I just had to post it for him =D


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## keemy (Dec 23, 2010)

I feel like 

\( .\overline{1}^{\displaystyle-\frac{6!!!\ldots}{2^n}} - .2^{\displaystyle-\frac{3!!\ldots}{2^m}} = 9^{\displaystyle\frac{6!!!\ldots}{2^n}} - 5^{\displaystyle\frac{3!!\ldots}{2^m}} \) (the 2^n and m coming from being able to sqrt) 

should have a lot of solutions between 1 and 100 (if not all) if anyone wants to look there.

edit: Also I was wondering if remainder/mod would be legal (as I recall learning remainder in elementary school).

edit 2: ya like -1 + 6! mod(.2^-3)) = 94


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## reThinking the Cube (Dec 23, 2010)

trying-to-speedcube... said:


> \( 44=.1^{-{\sqrt{\sqrt{\sqrt{\ldots{\sqrt{6}}}}}}}/.2-3! \)



Nice. I think the standard notation for that would probably be -

\( 44=. \)™\( 1^{-{\sqrt{\sqrt{\sqrt{\ldots{\sqrt{6}}}}}}}/.2-3! \)



Tyson said:


> Nice Keemy, it is less obscure than the double factorial no doubt. Unfortunately, this problem is given to sixth graders, and sixth graders actually don't learn normal factorial until after the New Year at this school. So really, to do the problem completely kosher, you're limited to the basic four operations +, -, *, /, decimals, repeating decimals, and factorials. Also powers, but you need a digit, and square root, but you don't need a digit. And you could use a cube root or something if you wanted, but you would need to utilize a digit, so perhaps you can use the 6th root. I wonder if that would help...
> 
> But yeah, obviously the more simple the solution, the more reasonable it is for a 6th grader to get




For 6th graders then, using any calculator the (**=) is equivalent to (^2), therefore:

44 = ((1+6)**)-2-3

EDIT: If this idea is acceptable, then 94 and 98 can be done with this trick too.


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## Carrot (Dec 23, 2010)

reThinking the Cube said:


> Nice. I think the standard notation for that would probably be -
> 
> \( 44=. \)™\( 1^{-{\sqrt{\sqrt{\sqrt{\ldots{\sqrt{6}}}}}}}/.2-3! \)


 
ehh... why would you ever put ™ inside a math statement? and then in between a dot and number?? And whatever you tried to tell me I didn't get it, because I don't know wether I should say "thanks" or "screw you" xD

ohh btw, Maarten wanted me to post this too:


Spoiler



JOKE SOLUTIONS!! xD

1...6*23 http://www.wolframalpha.com/input/?i=1...6*23
He found 1...6^2*3


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## maggot (Dec 23, 2010)

i also came up with that sqrt sqrt sqrt sqrt 6 soln, but its kinda ugly imho. but, after many hours, i dont think there is a pretty soln.


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## qqwref (Dec 23, 2010)

Hm, hang on...

\( 98 = -\sqrt{\sqrt{.\overline{1}^{-6}}} + .2^{-3} \)


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## Tyson (Dec 24, 2010)

qqwref said:


> Hm, hang on...
> 
> \( 98 = -\sqrt{\sqrt{.\overline{1}^{-6}}} + .2^{-3} \)


 
I bow to you.


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## HavoCentral (Dec 24, 2010)

Okay, I follow all of the math, but how/where do you get/make the math images from.


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## Stefan (Dec 24, 2010)

If you ever wonder how someone wrote something, just click "Reply With Quote" and look at it. And that's an awesome 98, btw.


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## vcuber13 (Dec 24, 2010)

HavoCentral said:


> Okay, I follow all of the math, but how/where do you get/make the math images from.


 
use [noparse]\( and \)[/noparse]
and then LaTeX as the code


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## maggot (Dec 24, 2010)

i like our fib soln better, but that is very elegant qq. . .


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