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BrainBoggler
Thou whoreson cullionly barbermonger


Joined: 08 Jun 2004
Posts: 120

PostPosted: Wed Aug 23, 2006 8:12 pm    Post subject: Reply with quote

Nice job bringing the challenge to a close, lotjegetikt! Thumbs up

Sean (BrainBoggler)
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VoyagerKing
O ho monster


Joined: 23 Aug 2006
Posts: 13
Location: Michigan

PostPosted: Wed Aug 23, 2006 8:33 pm    Post subject: FINAL SOLVED LIST Reply with quote

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!
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thesane
O ho monster


Joined: 19 Jun 2006
Posts: 22
Location: NZ

PostPosted: Wed Aug 23, 2006 8:40 pm    Post subject: Reply with quote

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

PostPosted: Thu Aug 24, 2006 3:31 pm    Post subject: Reply with quote

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. Razz
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Rebecca
Thou whoreson cullionly barbermonger


Joined: 16 Sep 2003
Posts: 128
Location: Perth

PostPosted: Thu Aug 24, 2006 5:37 pm    Post subject: Reply with quote

Hatfielder wrote:
but I'm obviously mean to my kids


I,m glad you weren't my teacher then - just kidding Angelic
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SuperGrover
Lets meet as little as we can


Joined: 12 Nov 2004
Posts: 1165
Location: USA

PostPosted: Thu Aug 24, 2006 7:07 pm    Post subject: Reply with quote

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. Razz


And I bet you will not let them use the Gamma function, either! Razz
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stozz
Thou knotty pated fool


Joined: 23 Mar 2004
Posts: 346
Location: Nottingham, England

PostPosted: Thu Aug 24, 2006 7:11 pm    Post subject: Reply with quote

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

PostPosted: Thu Aug 24, 2006 7:19 pm    Post subject: Reply with quote

I'd say all of them:

1 = 1
2 = 2
etc...

8)
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BrainBoggler
Thou whoreson cullionly barbermonger


Joined: 08 Jun 2004
Posts: 120

PostPosted: Thu Aug 24, 2006 7:27 pm    Post subject: Reply with quote

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^(6-2)

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

PostPosted: Thu Aug 24, 2006 7:29 pm    Post subject: Reply with quote

I'm guessing that that would only be allowed for 1-9 inclusive, with the question being interpreted as 'using only the individual digits, as numbers in their own right'. In which case, most 2-digit 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 Smile ) is:

25=5^2

Sad

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
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PostPosted: Thu Aug 24, 2006 7:33 pm    Post subject: Reply with quote

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. Razz


And I bet you will not let them use the Gamma function, either! Razz


Why not, you guys just did without asking Razz Wink
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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

PostPosted: Thu Aug 24, 2006 7:43 pm    Post subject: Reply with quote

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. Razz
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theandygrant
This is the foul fiend Flibbertigibbet


Joined: 07 Feb 2004
Posts: 1264
Location: The Lethargic Dodecahedron

PostPosted: Thu Aug 24, 2006 7:45 pm    Post subject: Reply with quote

What's the gamma function? Confused
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SuperGrover
Lets meet as little as we can


Joined: 12 Nov 2004
Posts: 1165
Location: USA

PostPosted: Thu Aug 24, 2006 8:09 pm    Post subject: Reply with quote

theandygrant wrote:
What's the gamma function? Confused


It's an important function in analysis and heavy-lifting-style probability theory. But on the integers, it is defined as gamma(n) = (n-1)! A handy-dandy feature is that gamma(1/2)= sqrt(pi).

In reality, I would be more likely to spring the odd-factorial function on my teacher: n!! = (n)(n-2)...(2) if n is even, (n)(n-2)...(3)(1) if n is odd.
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SuperGrover
Lets meet as little as we can


Joined: 12 Nov 2004
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PostPosted: Thu Aug 24, 2006 8:37 pm    Post subject: Reply with quote

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)*5-3)^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)
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