wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> hard >> HARD: 24 II
        
(Message started by: srowen on Jul 30^th, 2002, 6:19pm)

Title: HARD: 24 II
Post by srowen on Jul 30^th, 2002, 6:19pm

24 = (10-3)*2 + 10

Title: Re: HARD: 24 II
Post by icon on Jul 30^th, 2002, 6:37pm

i swore i just posted reply to this maybe 10 min ago but didnt even see you posting it before hehe

Title: Re: HARD: 24 II
Post by srowen on Jul 30^th, 2002, 6:56pm

Yeah, looks like we posted about 1 minute apart, how about that. What are the odds... sounds like another riddle in the making!

Title: Re: HARD: 24 II
Post by william wu on Jul 30^th, 2002, 7:15pm

on 07/30/02 at 18:37:25, 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.

Title: Re: HARD: 24 II
Post by icon on Jul 31^st, 2002, 3:56am

lol :) 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)

Title: Re: HARD: 24 II
Post by Eric Yeh on Aug 1^st, 2002, 6:59am

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

Title: Re: HARD: 24 II
Post by icon on Aug 1^st, 2002, 12:07pm

[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?

Title: Re: HARD: 24 II
Post by Eric Yeh on Aug 1^st, 2002, 2:41pm

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. :P

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

Title: Re: HARD: 24 II
Post by icon on Aug 2^nd, 2002, 5:45am

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:>

Title: Re: HARD: 24 II
Post by Eric Yeh on Aug 2^nd, 2002, 6:09am

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. :)

Best,
Eric

Title: Re: HARD: 24 II
Post by Brett Danaher on Aug 2^nd, 2002, 1:16pm

on 08/02/02 at 06:09:56, Eric Yeh wrote:

Anyone got any clues about 3 3 7 7? It's killing me... I really can't get this one to work out.

Title: Re: HARD: 24 II
Post by Eric Yeh on Aug 2^nd, 2002, 1:21pm

If you really really want one...

Think fractions.

Title: HARD: 24 II
Post by Brett Danaher on Aug 2^nd, 2002, 1:32pm

on 08/02/02 at 13:21:01, 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. :)

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.

Title: Re: HARD: 24 II
Post by Eric Yeh on Aug 2^nd, 2002, 1:36pm

Er... 4!-4-(4/4) ??

Am I missing something or am I just smoking today? ;) ;) ;)

Title: Re: HARD: 24 II
Post by icon on Aug 3^rd, 2002, 7:08pm

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

Title: Re: HARD: 24 II
Post by Eric Yeh on Aug 3^rd, 2002, 9:26pm

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

Title: Re: HARD: 24 II
Post by icon on Aug 3^rd, 2002, 10:57pm

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 :>

Title: Re: HARD: 24 II
Post by Eric Yeh on Aug 5^th, 2002, 8:38am

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? :)

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

Title: Re: HARD: 24 II
Post by icon on Aug 5^th, 2002, 10:37am

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

Title: Re: HARD: 24 II
Post by Brett Danaher on Aug 7^th, 2002, 8:55am

on 08/02/02 at 13:36:37, Eric Yeh wrote:

Er... 4!-4-(4/4) ??

Am I missing something or am I just smoking today? ;) ;) ;)

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.

Title: Re: HARD: 24 II
Post by NickH on Aug 7^th, 2002, 2:55pm

"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

Title: Re: HARD: 24 II
Post by scodger on Sep 12^th, 2002, 5:30am

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

Title: Re: HARD: 24 II
Post by Eric Yeh on Sep 12^th, 2002, 6:02am

5*(5-1/5)