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#26 Re: Puzzles and Games » Hidden countries » 2012-05-03 10:09:04
never heard of it.
like I say I can't find any mention of another country of that letter.
#27 Re: Puzzles and Games » Hidden countries » 2012-05-03 09:59:24
Again,it depends on the source. Google for countries that start with that letter. It should be easier.
just done that and every site I've tried says the same thing that there's only one, Are we talking about the same country here?
#28 Re: Puzzles and Games » Hidden countries » 2012-05-03 09:47:16
It is not a real country. It depends on the source.
OK but you still said there were other countries beginning with that letter, yet according to that list I just saw there aren't any? I don't understand.
#29 Re: Puzzles and Games » Hidden countries » 2012-05-03 09:41:54
The one they wrote isn't considered an independent country,but there are others that begin with that letter.
what do you mean?
I just looked at a list of all the countries in the world and first that country was listed and second there were no other countries of that letter!
#30 Re: Puzzles and Games » Hidden countries » 2012-05-03 09:20:53
I got 10/10 no help from anyone or anything!:D
the 2nd one was a bit tricky but I got there in the end, the last one was a bit tough (useless fact: it's the only country that begins with that letter!!!)
what I looked for was unusually word combinations:D
#31 Re: Puzzles and Games » Perfect Squares » 2012-04-16 03:59:35
Here is my solution: It is based to the fact that each perfect square N^2 is the sum of the first N odd numbers (5^2 = 25 = 1+3+5+7+9).
Thus the difference of any 2 perfect squares should equal to the sum of consecutive odd numbers (and this should equal 60).
Starting from 1, we write down the sums of the odd numbers:
1+3+5+7+9+11+13 = 49 while 1+3+5+7+9+11+13+15 = 64. Thus we cannot make 60 starting from 1.
We do the same with 3: 3+5+7+9+11+13=48 while 3+5+7+9+11+13+15 = 63. Not possible.
Starting from 5:
5+7+9+11+13+15=60 Here we are.
So, one perfect square is 4 and the next is 64, their difference being 60, so the first number we are looking for is 34 (34-30 = 4 and 34+30 = 64, both of them perfect squares).
Similarly, we find that the only other sum of consecutive odd numbers equaling 60 is 29+31.
Therefore one perfect square is 1+3+5+...+27=196 (14^2) and the next is 1+3+5+...+27+29+31=256 (16^2) and the second number we are asking for is 226 (226-30 = 196, 226+30 = 256).
There is no other series of successive odd numbers equaling 60, so these two are the only numbers with this property.
Obviously this is not a proper "proof"; it is more based on a "guess and try" method, but it works!
Have you seen my proof? It's very similar in that it's based on consecutive odd numbers.
#32 Re: Puzzles and Games » Perfect Squares » 2012-04-11 21:39:11
And I did it non-experimentally!
I did it diferent again:D
#33 Re: Puzzles and Games » The Professors Test » 2012-03-29 08:11:43
I thought this might be better in a new post
#34 Re: Puzzles and Games » The Professors Test » 2012-03-29 06:35:15
hi wintersolstice and anonimnystefy
Did you mean:
A says None of us is lying. B says Only two of us are lying. C says Only one of us is lying. D says all of us are lying.
The truth table below shows there is no solution to this.
The middle code must have 84 characters if it is to work. A quick count suggests there are less. But some spaces are double. Your problem is to work out which.
Bob
I also must note that if we add the third person saying that they are all lying then the puzzle turns into a paradox.
I see what you mean
I figured out where the paradox is now
I think I need to re-do the liars puzzle completely. Sorry about that.
#35 Re: Puzzles and Games » The Professors Test » 2012-03-28 09:15:43
hi wintersoltice
have you solve any other problems?
yes I solved a few of them but instead of posting my answers I checked them against yours (I haven't solved all the ones that you solved yet!:D)
I also must note that if we add the third person saying that they are all lying then the puzzle turns into a paradox.
how come? plus I said 4th person:D
#36 Re: Puzzles and Games » The Professors Test » 2012-03-27 08:11:51
hi wintersolstice
That's exactly what I had and I also was surprised by anonimnystefy's answer. Under the circumstances of the test, being surprised is entirely appropriate.
Bob
#37 Re: Puzzles and Games » The Professors Test » 2012-03-27 02:48:53
hi Stefy,
Good start.
Bob
#38 Re: Puzzles and Games » The Professor and the Student » 2012-03-21 07:12:07
hi
Of course, I agree with amberzak!
But for reasons of logic too.
The deduction "We've reached the end of Thursday, so it must be on Friday" is sound.
but this does not allow you to deduce anything about earlier days.
Because, for example, at the end of Wednesday the logic is "We've reached the end of Wednesday, so it could be on Thursday or Friday". It is always dangerous to keep applying the same logic as the student has done. And MathsIsFun's answer shows why!
Bob
I don't actually agree with this logic because if they reached the end of Wednesday then it must be Thursday because if it was on Friday it wouldn't be a suprise (because if it hasn't happen by the end of Thursday is has to be Friday)
but that's just my opinion:D I have seen another way (apart from my own) of resolving th paradox so I suppose it show there's more than one way to do it:D
#39 Re: Puzzles and Games » The Professor and the Student » 2012-03-20 03:39:11
the way I resolve this paradox is to say that:
the students use the logic to eliminate every possible day, so whichever day it happens it's a a suprise (they didn't see it coming) because they eliminated every possible day:D
#40 Re: Puzzles and Games » Word Game 23c » 2012-03-18 05:44:35
here's my submission (just one word:D)
#41 Re: Puzzles and Games » Word Game #66 » 2012-03-17 09:23:09
if this is still on:D
#42 Puzzles and Games » Variation of "the hardest logic puzzle" » 2012-03-17 06:36:10
- wintersolstice
- Replies: 1
I've got a bit of an interesting variation to The Hardest Logic Puzzle EVER"
(one of a few potential varitions I have hought about)
The differences are
- You have a new Chief (red) who won't answer any question (he can speak but he won't speak to you so you can't ask him)
- one of the chiefs always answers "Yes" (in their own language of course) regardless of what the question is!
- you goal is to find this chief ONLY. (he may be the red chief he may not be)
so assuming all the other rules are the same (check the link if you need to:D)
find 3 question you can ask.:D
#43 Re: This is Cool » Super Happy Numbers » 2012-03-05 01:08:52
that new axiom sounds better:D
btw why are you thanking for the proof I haven't posted it yet?:D
I only posted the statement that I'd proved:m
when I post the proof I'll also prove that the rules are the same for "Happy numbers, Super happy numbers, Hyper 3 happy numbers etc
I'm also looking at higher powers, other bases (binary, ternary etc) and like I've already said other intrval sizes:D
It's become an obsession atm!
#44 Maths Is Fun - Suggestions and Comments » problem with smilie » 2012-03-04 08:58:57
- wintersolstice
- Replies: 4
usully I click on the 
smilie it remains (as letters and symbols) I'm not sure how amny other smilies don't work
here a link to a thread where I used it and it didn't work
http://www.mathisfunforum.com/viewtopic.php?id=17315&p=2
#45 Re: This is Cool » Super Happy Numbers » 2012-03-04 08:49:03
Hi wintersolstice,
"the largest number with 3 intervals"
For SHNs, that probably only needs to be "the largest number with 2 intervals" - depending on your proof wording (ie, if "largest number" refers to the test number only and not any of its iterations: eg, 9999 iterates to 19602, which has 3 intervals).
Yes...I'd like to see those proofs. Always interested to see how others do things...might be handy for something later.
well actually my proof proves the following statement
whenimplying that a number with 4 or more intervals will always have less intervals after an iteration and that a number with 3 intervals can have up to 3 intervals (so it may have an equal value after an iteration)
this means your proof has a slight hole in it:
there could be a number between 9999 and 999999 (i.e 3 intervals) that is "equal" after an iteration
you haven't proven/disproven that yet
such a number could form a cycle on it's own!
so you still need to check up to 999999 just in case (though I doubt such a number exists, it's best to be on the safe side )
#46 Re: This is Cool » Super Happy Numbers » 2012-03-03 08:16:04
Hi wintersolstice,
You don't have to check numbers >9999 because that is the largest number that produces a number greater than itself in its first iteration...which means that numbers higher than that only lead to re-checking numbers already checked.
well what I've done is to demonstrate that if you search between one and "the largest number with 3 intervals" you can always find the entire list of cycles for any type of Hyper-n happy numbers (although as you've demonstrated it might not be necessary to search them all:D)
I discovered that property when first working on these cycles, but forgot to tell you.
no worries:D I had a lot of fun with all that complicated proving (I'm now going to take it to other bases and other powers and to all Hyper-n cases:D)
do you think it's worth still posting the proofs? (because it's strange how those rules work with all interval sizes)
A test I did shows that any number will eventually iterate to one of the twelve cycles that I mentioned in post #6, so I reckon we won't find any others.
well it's looks like for super happy numbers the problem is solved:D
#47 Re: This is Cool » Super Happy Numbers » 2012-03-02 07:59:54
if you can exhaust all numbers less than a million you'll find all possible loops this is because
when put into the procedure:
a 1 interval number will become a 1 or 2 interval number
a 2 interval number will become a 1,2 or 3 interval number
a 3 interval number will become a 1,2 or 3 interval number
a n interval number will become a "less than n" interval number (where n is 4 or more)
so any number will fall below 4 intervals and then stay there
these rules are universal
meaning they work regardless of what the interval size is!
in other words they work for
happy numbers
super happy numbers
hyper-n happy numbers (regardless of what n is!)
when I can I'll post a complete proof of all of this:D
#48 Re: This is Cool » New Math Formula: Sums of Power for Arithmetic Series » 2012-02-04 08:00:55
I wouldn't mind knowing how you figured this out/found it because I was working on a formula using a completly different method:
two pyramids of numbers generated using Pascals Triangle, then the use of simaltaneous equations
this worked out the indivdual coffiecents of the terms of the formula all you have to do is factorise.
or was it the same method?
#49 Re: This is Cool » Super Happy Numbers » 2012-02-02 08:41:58
well I'll try and find some info on "JustBasic" (never heard of it!LOL) it doesn't matter that you can't help very much:D
btw a "n-Hyper Happy number" is a generalisation I invented just after the "Super Happy numbers" basically you break the number down into intervals of "n" (normal happy numbers n=1, super happy numbers n=2) you would need to add zeros to the beginning of a number (if necessary) to make sure it has a number of digits which is a multiple of "n" the same rules apply:D
#50 Re: This is Cool » Super Happy Numbers » 2012-02-02 01:38:42
Hi wintersolstice
btw what method are you using how did you work them out so fast?
I wrote a program in BASIC using the LibertyBASIC software program, testing all numbers below 100,000 against the rules.
Running time to find those 244 numbers was about 90 seconds, which would be pretty slow compared to other languages and programs like Mathematica - none of which I know.
I'm sure a good BASIC programmer would do better (I just potter around), and running it through a compiler would improve the speed too.
well I know absolutly nothing about programming whatsoever:(:( but have longed to learn how to do it. this is just one of hundreds of things I can't do because I don't even know the basics of pogramming:(
can you offer any help? (anything would be appreciated:D)
I had the idea of extending to "n-Hyper Happy numbers" aswell