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   HARD: 24 II
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   Author  Topic: HARD: 24 II  (Read 2715 times)
S. Owen
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HARD: 24 II  
« on: Jul 30th, 2002, 6:19pm »
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24 = (10-3)*2 + 10
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Re: HARD: 24 II  
« Reply #1 on: Jul 30th, 2002, 6:37pm »
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i swore i just posted reply to this maybe 10 min ago but didnt even see you posting it before hehe
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S. Owen
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Re: HARD: 24 II  
« Reply #2 on: Jul 30th, 2002, 6:56pm »
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Yeah, looks like we posted about 1 minute apart, how about that. What are the odds... sounds like another riddle in the making!
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william wu
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Re: HARD: 24 II  
« Reply #3 on: Jul 30th, 2002, 7:15pm »
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on Jul 30th, 2002, 6:37pm, icon wrote:
i swore i just posted reply to this maybe 10 min ago but didnt even see you posting it before hehe

 
yea, i deleted icon's post because it was redundant. you guys were less than a minute apart.
 
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Re: HARD: 24 II  
« Reply #4 on: Jul 31st, 2002, 3:56am »
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lol Smiley my old teacher used to do this 24 ones except he made 10 or so and timed us on who gets it fastert(all 10)
 
i have to admin this1 took me maybe 1-2 min max without pen or paper but other 1 i spend few hrs on it hehe(didnt realise about fractions)
 
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Eric Yeh
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Re: HARD: 24 II  
« Reply #5 on: Aug 1st, 2002, 6:59am »
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Ye, when I offhandedly mentioned this one to Will I by no means meant it as a difficult one to get -- it was more in the context of having an interesting solution.
 
Hmm, maybe this makes it a better problem, if we are trying to make it fit the "hard" category:  Expand to a full operator set +-*/^ (this is the way I usually play 24, btw Will -- youll note it gives a lot more interesting solutions in general to otherwise unsolvables like 1 1 2 5 (as an easy example)).
 
Find all solutions.  (Equivalently, how many distinct solns are there?  I could define distinct for you guys but I'll leave it as an exercise to the reader -- there are some subtleties that can be matter of choice, but it is a matter of self-satisfaction.)
 
Happy puzzling,
Eric
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Re: HARD: 24 II  
« Reply #6 on: Aug 1st, 2002, 12:07pm »
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[quote author=Eric Yeh link=board=riddles_hard;num=1028078355;start=0#5 date=08/01/02 at 06:59:30]
 
Hmm, maybe this makes it a better problem, if we are trying to make it fit the "hard" category: šExpand to a full operator set +-*/^ (this is the way I usually play 24, btw Will -- youll note it gives a lot more interesting solutions in general to otherwise unsolvables like 1 1 2 5 (as an easy example)).
 
do you mean alloweing square/square root of, powers/factorials?
 
is no then its like (5 to square root of 2 -1) *1
 
or am i confused?
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Eric Yeh
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Re: HARD: 24 II  
« Reply #7 on: Aug 1st, 2002, 2:41pm »
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I mean to allow exponentiation by itself (^).
 
I suppose one could also allow roots, since it would be parallel to subtraction and division, but for some reason I find this cheesy and like to require an additional 1 to get this behavior (e.g. you need 1, 2, 9 to get 9^(1/2) = 3).  At that point you could add log as well.   Tongue
 
But it's all personal taste.
 
As far as your example, I think you mean to leave out the "root" part otherwise you're right.
 
Best,
Eric
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Re: HARD: 24 II  
« Reply #8 on: Aug 2nd, 2002, 5:45am »
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hi
 
well actually when we used to doit, it was like it is now but adding roots/etc would just add 2-3 + answers for each 24
 
so its a matter of preference
 
so example the 3 3 7 7 and 1 3 4 6 answers are very elegant and unless u ever tried to solve this might take u quite a while to figure it out:>
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Eric Yeh
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Re: HARD: 24 II  
« Reply #9 on: Aug 2nd, 2002, 6:09am »
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Icon,
 
I completely agree that 3 3 7 7 and 1 3 4 6 are supremely elegant.  Although the addition of ^ does give an easy solution to the latter, you'll at least be happy to know that it leaves the beauty of 3 3 7 7 unadulterated.   Smiley
 
Best,
Eric
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Re: HARD: 24 II  
« Reply #10 on: Aug 2nd, 2002, 1:16pm »
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on Aug 2nd, 2002, 6:09am, Eric Yeh wrote:
Icon,
 
I completely agree that 3 3 7 7 and 1 3 4 6 are supremely elegant.  Although the addition of ^ does give an easy solution to the latter, you'll at least be happy to know that it leaves the beauty of 3 3 7 7 unadulterated.   Smiley
 
Best,
Eric

 
 
Anyone got any clues about 3 3 7 7?  It's killing me... I really can't get this one to work out.
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Eric Yeh
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Re: HARD: 24 II  
« Reply #11 on: Aug 2nd, 2002, 1:21pm »
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If you really really want one...
 
Think fractions.
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HARD: 24 II  
« Reply #12 on: Aug 2nd, 2002, 1:32pm »
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on Aug 2nd, 2002, 1:21pm, Eric Yeh wrote:
If you really really want one...
 
Think fractions.

 
Before I even saw your message, I started thinking about fractions and had it within 1 minute.  I can't believe I didn't think of that before.
 
(3+3/7)*7
 
I'm a moron.  Smiley  
 
Anyway, we used to play a similar game.  Given 4 4 4 4 and some set of operators, what numbers can you get?  We tried to go from 1 to 100, allowing +,-,/,* and also the root symbol (which, without a number gives you the square root or with the number n gives you the nth root).  We also allowed powers, factorial, and decimal points.  That made it too easy.  However, I challenge you this (and I can't find a solution myself).  Without using decimal points, use 4 4 4 4 and the above operators (even including factorial) to get 19.  I can't do it.  The only solution I ever came up with was (4 + sqrt 4) / .4 + 4.  That's 6/.4 = 15.  15 +4 = 19.  Now do it without a decimal.
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Eric Yeh
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Re: HARD: 24 II  
« Reply #13 on: Aug 2nd, 2002, 1:36pm »
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Er...  4!-4-(4/4) ??
 
Am I missing something or am I just smoking today?   Wink   Wink   Wink
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Re: HARD: 24 II  
« Reply #14 on: Aug 3rd, 2002, 7:08pm »
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actually good 24 questions are rare and the key involves not only in having a fraction but also like having a hard way to figure it out
 
like
 
1 3 4 6 is a good example, took me about a day to doit cause its a tricky even when u figure 1 part out
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Eric Yeh
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Re: HARD: 24 II  
« Reply #15 on: Aug 3rd, 2002, 9:26pm »
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Icon,
 
I'm not sure I understand -- what "1 part" is that?  It seems to me that there's just one major "innovation" in the solution to 1 3 4 6:  using the fraction to divide out.   Is there another way to think about it that splits it into two "parts"?
 
Best,
Eric
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Re: HARD: 24 II  
« Reply #16 on: Aug 3rd, 2002, 10:57pm »
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well u see
 
when i solved 24 i couldnt really sit at pc or with paper or with anything
 
i just was actually sitting in my plastic chair and getitng some sun tan and thinming
 
2nd key thing for me was 2nd / like u get
 
 6 / (
 
i kinda was slow to figure when u do this via fractions u switch it
 
anyhow it was my own issue hehe probably too much time in the sun :>
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Eric Yeh
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Re: HARD: 24 II  
« Reply #17 on: Aug 5th, 2002, 8:38am »
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Hold on, then what was the first key?  Don't tell me you got the 1-3/4 and were sitting on that for a while before seeing how to combine 6 and .25?   Smiley
 
Sorry, I don't mean to just harp on this -- I am curious to see if there's another way of thinking that can give you a soln like this in stages.  I can't think of how one could get this problem any way but all at once.
 
Best,
Eric
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Re: HARD: 24 II  
« Reply #18 on: Aug 5th, 2002, 10:37am »
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hi
 
not that really i was considering fractions at 1st but what took me a while was the fact that 6 x .25 is same as 6  x 4
 
<---wierdo hehe
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Re: HARD: 24 II  
« Reply #19 on: Aug 7th, 2002, 8:55am »
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on Aug 2nd, 2002, 1:36pm, Eric Yeh wrote:
Er...  4!-4-(4/4) ??
 
Am I missing something or am I just smoking today?   Wink   Wink   Wink

 
Sorry - I forgot.  The challenge we gave ourselves was to do it without factorial and without the use of decimal points.  So you have exponents and roots and the 4 basic ops, and four 4's.  You have to get 19.  We believed it could not be done.  I'd love to be proven wrong.
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Re: HARD: 24 II  
« Reply #20 on: Aug 7th, 2002, 2:55pm »
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"Anyway, we used to play a similar game.  Given 4 4 4 4 and some set of operators, what numbers can you get?  We tried to go from 1 to 100, allowing +,-,/,* and also the root symbol (which, without a number gives you the square root or with the number n gives you the nth root)."
 
If you also allowed log, you could obtain any positive integer as follows:
 
n = -log4 (log4 (root...root(4)))
 
where there are 2n nested radical symbols.
 
I'm afraid I can't claim the credit for this!  Someone, I forget who -- may have been von Neumann -- immediately gave n = -log2 (log2 (root...root(2))) as the basis of a general solution, when confronted with the similar four 2's question.  It may be apochryphal, but it's a great story!
 
Nick
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Re: HARD: 24 II  
« Reply #21 on: Sep 12th, 2002, 5:30am »
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how about 5 5 5 1
therre were cards like this you could buy, and this was the one on the back of the box, called the ultimate challenge
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Eric Yeh
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Re: HARD: 24 II  
« Reply #22 on: Sep 12th, 2002, 6:02am »
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5*(5-1/5)
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