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BrainBoggler Thou whoreson cullionly barbermonger
Joined: 08 Jun 2004 Posts: 120

Posted: Wed Aug 23, 2006 8:12 pm Post subject: 


Nice job bringing the challenge to a close, lotjegetikt!
Sean (BrainBoggler) 

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VoyagerKing O ho monster
Joined: 23 Aug 2006 Posts: 13 Location: Michigan

Posted: Wed Aug 23, 2006 8:33 pm Post subject: FINAL SOLVED LIST 


Man,
You're done!
You've just posted expressions for the final 3 integers, 39,41, and 55, and that means you're done. Very Nicely Done!
Just one little item I'd like to mention, you were not allowed to introduce any additional numerical digits, so I *removed* the zeroes from the final 3 expressions, but of course, that did not change how they calculated out.
Thanks a lot for running my challenge through your mental mill. I hope you had fun.
Here then, is The Final Solved List:
1 = (4+4)/(4+4)
2 = (4/4)+(4/4)
3 = (4+4+4)/4
4 = sqrt(4*4)*(4/4)
5 = (4x4 + 4) / 4
6 = (4! / 4) + 4  4
7 = (4+4)  (4/4)
8 = 4 + 4 + 4  4
9 = 4/.4  4/4
10 = 4/.4 + 4 – 4
11 = (4! / sqrt(4))  (4 / 4)
12 = (4 x 4)  sqrt(4)  sqrt(4)
13 = (4! / sqrt(4)) + (4 / 4)
14 = (4 x 4)  4 + sqrt(4)
15 = (4 x 4)  (4 / 4)
16 = 4 + 4 + 4 + 4
17 = (4 x 4) + (4 / 4)
18 = (4! / sqrt(4)) + 4 + sqrt(4)
19 = 4!  4  (4 / 4)
20 = ((4! x 4) / 4) – 4
21 = 4!  4 + (4 / 4)
22 = (4 x 4) + 4 + sqrt(4)
23 = 4!  ((4 / 4) ^ 4)
24 = 4!  4 + sqrt(4) + sqrt(4)
25 = 4! + ((4 / 4) ^ 4)
26 = 4! + sqrt(4) + 4 – 4
27 = 4! + sqrt(4) + (4 / 4)
28 = 4! + 4 + 4  4
29 = 4! + 4 + (4 / 4)
30 = 4! + 4 + 4  sqrt(4)
31 = ((4! / .4) + sqrt(4)) / sqrt(4)
32 = 4! + sqrt(4) + sqrt(4) + 4
33 = 4! + 4 + (sqrt(4) / .4)
34 = 4! + (4! / sqrt(4))  sqrt(4)
35 = ((4 * 4)  sqrt(4)) / .4
36 = 4! + 4 + 4 + 4
37 = ((Sqrt(4) + 4!) / Sqrt(4)) + 4!
38 = 4/.4 x 4  sqrt(4)
39 = ((4 + Sqrt(4)) / .4) + 4!
40 = 4! + (4 + 4) x sqrt(4)
41 = ((Sqrt(4) + 4!) / .4)  4!
42 = (4! * sqrt(4))  4  sqrt(4)
43 = (4! * sqrt(4))  (sqrt(4) / .4)
44 = ((4 * 4) / .4) + 4
45 = ((4 * 4) + sqrt(4)) / .4
46 = (4! * sqrt(4))  (4 / sqrt(4))
47 = 4! x sqrt(4)  (4/4)
48 = 4! x sqrt(4) x (4/4)
49 = 4! x sqrt(4) + (4/4)
50 = 4!/.4  4/.4
51 = (4!sqrt(4))/.4 – 4
52 = 4! + 4! + sqrt(4) + sqrt(4)
53 = (4!sqrt(4))/.4  sqrt(4)
54 = (4!4)/.4 + 4
55 = ((4!  .4) / .4)  4
56 = 4! + 4! + 4 + 4
57 = (4!sqrt(4))/.4 + sqrt(4)
58 = 4!/.4  4/SQRT(4)
59 = 4!/.4  4/4
60 = [(4x4) x 4] – 4
Over and Out! _________________ VoyagerKing 

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thesane O ho monster
Joined: 19 Jun 2006 Posts: 22 Location: NZ

Posted: Wed Aug 23, 2006 8:40 pm Post subject: 


Just to throw a new spin onto this puzzle, I have seen this sort of thing before (although not this exact version), in which you use the digits of the current year (and try to get to 100). Admittedly, since 2000 this has probably become quite an interesting prospect, I originally got given the problem around 1993. I can't remember if I ever completed a year, but I had a go at them all at the time (and mostly gave up in 2000). 

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Hatfielder A great lubbery boy
Joined: 19 May 2004 Posts: 233 Location: The beautiful city of Durham, well a nearby village anyway

Posted: Thu Aug 24, 2006 3:31 pm Post subject: 


I'm quite amazed that you haven't seen the four fours problem before. I remember doing it as a homework task when I was 12 (2nd year high school). I've used it a few times with some of my classes  but I'm obviously mean to my kids  I won't let them use .4 and I get them to try to make all the numbers up to 100. 

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Rebecca Thou whoreson cullionly barbermonger
Joined: 16 Sep 2003 Posts: 128 Location: Perth

Posted: Thu Aug 24, 2006 5:37 pm Post subject: 


Hatfielder wrote:  but I'm obviously mean to my kids 
I,m glad you weren't my teacher then  just kidding _________________ Growing old is INEVITABLE. Growing up is OPTIONAL 

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SuperGrover Lets meet as little as we can
Joined: 12 Nov 2004 Posts: 1165 Location: USA

Posted: Thu Aug 24, 2006 7:07 pm Post subject: 


Hatfielder wrote:  I've used it a few times with some of my classes  but I'm obviously mean to my kids  I won't let them use .4 and I get them to try to make all the numbers up to 100. 
And I bet you will not let them use the Gamma function, either! _________________ There are only 10 kinds of people in the worldthose who know binary, and those who don't. 

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stozz Thou knotty pated fool
Joined: 23 Mar 2004 Posts: 346 Location: Nottingham, England

Posted: Thu Aug 24, 2006 7:11 pm Post subject: 


A puzzle along the same lines for the latent (and not so latent!) mathematicians:
How many numbers can you find which can be mathematically expressed using their own digits?
For example (3+4)^3 = 343.
Jon 

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lotjegetikt Thou odoriferous stench
Joined: 18 Jul 2004 Posts: 1733 Location: The Netherlands

Posted: Thu Aug 24, 2006 7:19 pm Post subject: 


I'd say all of them:
1 = 1
2 = 2
etc...
8) _________________ The Crackerjacks 

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BrainBoggler Thou whoreson cullionly barbermonger
Joined: 08 Jun 2004 Posts: 120

Posted: Thu Aug 24, 2006 7:27 pm Post subject: 


Jon, If the digits don't have to be in the same order as the actual number, then these would work:
121 = 11^2
625 = 5^(62)
I imagine keeping the numbers in the same order increases the challenge.
Sean (BrainBoggler) 

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Robin Scurvy, old, filthy, scurry lord
Joined: 06 Mar 2004 Posts: 501 Location: Durham, England

Posted: Thu Aug 24, 2006 7:29 pm Post subject: 


I'm guessing that that would only be allowed for 19 inclusive, with the question being interpreted as 'using only the individual digits, as numbers in their own right'. In which case, most 2digit numbers, at least, seem to be completely impossible, at least using only the basic 4 operations, along with powers. In fact, the only one I can come up with (although I've not checked exhaustively, so I may have missed 1 or 2 ) is:
25=5^2
With more digits, of course, you have much more flexibility, so there would, I guess, be a reasonable, if modest, number of those. And if you start allowing factorials etc., then you can probably do a bit better. 

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Robin Scurvy, old, filthy, scurry lord
Joined: 06 Mar 2004 Posts: 501 Location: Durham, England

Posted: Thu Aug 24, 2006 7:33 pm Post subject: 


SuperGrover wrote:  Hatfielder wrote:  I've used it a few times with some of my classes  but I'm obviously mean to my kids  I won't let them use .4 and I get them to try to make all the numbers up to 100. 
And I bet you will not let them use the Gamma function, either! 
Why not, you guys just did without asking 

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Hatfielder A great lubbery boy
Joined: 19 May 2004 Posts: 233 Location: The beautiful city of Durham, well a nearby village anyway

Posted: Thu Aug 24, 2006 7:43 pm Post subject: 


Any of them who use the Gamma function get extra homework for cheating and just getting the answers from one of the many hundreds of webpages on this problem. 

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theandygrant This is the foul fiend Flibbertigibbet
Joined: 07 Feb 2004 Posts: 1264 Location: The Lethargic Dodecahedron

Posted: Thu Aug 24, 2006 7:45 pm Post subject: 


What's the gamma function? 

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SuperGrover Lets meet as little as we can
Joined: 12 Nov 2004 Posts: 1165 Location: USA

Posted: Thu Aug 24, 2006 8:09 pm Post subject: 


theandygrant wrote:  What's the gamma function? 
It's an important function in analysis and heavyliftingstyle probability theory. But on the integers, it is defined as gamma(n) = (n1)! A handydandy feature is that gamma(1/2)= sqrt(pi).
In reality, I would be more likely to spring the oddfactorial function on my teacher: n!! = (n)(n2)...(2) if n is even, (n)(n2)...(3)(1) if n is odd. _________________ There are only 10 kinds of people in the worldthose who know binary, and those who don't. 

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SuperGrover Lets meet as little as we can
Joined: 12 Nov 2004 Posts: 1165 Location: USA

Posted: Thu Aug 24, 2006 8:37 pm Post subject: 


stozz wrote:  A puzzle along the same lines for the latent (and not so latent!) mathematicians:
How many numbers can you find which can be mathematically expressed using their own digits?
For example (3+4)^3 = 343.
Jon 
If we have to use the digits in order, I submit:
729 = (7+2)^sqrt(9)
6859 = (6+8+5)^sqrt(9)
59319 = (5*9  3!*1)^sqrt(9)
117649 = ((1*1*7!/6!)^sqrt(4))^sqrt(9)
185193=((1+8)*(5+1)+sqrt(9))^3
205379 = ((2!+0!)+(5+3)*7)^sqrt(9)
456533 = ((sqrt(4)*5+6)*53)^3
493039 = (4!*sqrt(9)+3^0+3!)^sqrt(9)
658503 = ((6+5)*(8)5^0)^3
970299 = ((9+(7*0)+2)*9)^sqrt(9) _________________ There are only 10 kinds of people in the worldthose who know binary, and those who don't. 

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